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Lecture Comments (6)

1 answer

Last reply by: Professor Selhorst-Jones
Mon Mar 24, 2014 10:09 AM

Post by Sarang Janakiraman on March 23, 2014

On # 3 of Example 1, can we also say the limit is -∞? That's how I learnt it but would you find that acceptable?

1 answer

Last reply by: Professor Selhorst-Jones
Mon Nov 4, 2013 3:26 PM

Post by edick safarians on November 4, 2013

If you were to graph these functions, what feature would appear in the graph at the limiting value? Why?

1 answer

Last reply by: Professor Selhorst-Jones
Sun Jul 28, 2013 8:28 PM

Post by Jason Todd on July 27, 2013

Do you have any videos for Calculus, Differential equations, and/or Linear algebra somewhere else? I'm reviewing all of my math for engineering and your videos are a God-send.

Limits at Infinity & Limits of Sequences

  • Important Idea: infinity is not a location. It is the idea of going on forever, moving on to ever larger numbers. You can travel towards ∞, but you can never reach ∞.
  • We denote the limit of a function at infinity with

    lim
     
     f(x)               and              
    lim
     
     f(x).
    They mean the value that f(x) approaches as x goes off toward −∞ or ∞, respectively.
  • A limit at infinity works very similarly to how a normal limit works. Does the function "settle down" to a specific value L? It's just in this case, we're asking about the long-term behavior instead of x→ c. [It is important to note that most of the functions we're used to working with do not have limits at infinity.]
  • The type of functions we will work with most often that can have a limit at infinity are rational functions. We worked with these many lessons ago when we learned about asymptotes. They are functions created from a fraction with polynomials in the numerator and denominator:
    f(x) = an xn + an−1 xn−1 + …+ a1 x + a0

    bm xm + bm−1 xm−1 + …+ b1 x + b0


     
    .
    Notice that n is the numerator's degree and m is the denominator's degree. Then we can find limits at infinity by the below:
    • If n < m     ⇒     limx→ ±∞ f(x)  = 0
    • If n=m     ⇒     limx→ ±∞ f(x)  = [(an)/(bm)]
    • If n > m     ⇒     limx→ ±∞ f(x)  does not exist
  • In general, when trying to find limits at infinity, think in terms of how the function will be affected as x grows very large (positive and negative). Does the function grow without bound? Will it "settle down" over time? Two good ways to think about this are:
    • What happens if we plug in a LARGE NUMBER?
    • What are the rates of growth in the function? Which parts grow faster? Which parts grow slower?
    By thinking through these questions, you can get a good idea of how the function will behave over the long-term.
  • We can take this idea of a limit going to infinity and apply it to a sequence as well. Let a1, a2, a3,...   be some infinite sequence. Then we can consider

    lim
     
     an.
    The limit of a sequence is very similar to the limit of a function at infinity. Does the sequence "settle down" to a specific value L as n→ ∞?
  • Sometimes it can be difficult to tell how a function or sequence will behave in the long-run. In that case, we can evaluate the function numerically: plug in numbers and see what comes out. By using a calculator, we can plug in very large numbers (positive and negative) and see what happens to the function or sequence. Doing this will give us a good sense of the long-term behavior. Similarly, if you have access to a graphing calculator or program, you can graph the function. Expand the viewing window to a large horizontal region and look to see if the graph "settles down" in the long-run.

Limits at Infinity & Limits of Sequences

Complete the table below to approximate the limit at infinity.

lim
x→ ∞ 
  27x

3x2−5x
      
x
10
100
1000
10 000
100 000
f(x)
             
             
             
             
             
  • The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinity-that is, as x gets very, very large. [Notice that not all functions will have a limit at infinity.]
  • One way to find the value of this limit (or if it even exists) is by using larger and larger values for x. If plugging in really big values for x makes the function "settle down" towards a specific value, then we have found the value of the limit. If, on the other hand, larger and larger values give significantly different numbers each time, then the limit does not exist.
  • For this problem, we're finding the value of the limit by trying out some very large numbers. We plug in various large numbers, using an even bigger one each time. To complete the table, just plug each number in to the function, then use a calculator to get an approximate value. For example, the first two x-values would work as follows:
    x=10     ⇒     27(10)

    3(10)2 −5(10)
        ≈     1.080  000

    x=100     ⇒     27(100)

    3(100)2 −5(100)
        ≈     0.091 525
  • Continuing this process with each of the x-values, the completed table is
    x
    10
    100
    1000
    10 000
    100 000
    f(x)
    1.080  000
    0.091 525
    0.009 015
    0.000  900
    0.000 090
    Thus, looking at the table, we see that as x→ ∞, the value of the function is shrinking to extremely tiny numbers. As x→ ∞, we see f(x) → 0. Therefore our infinite limit is   limx→ ∞  [27x/(3x2−5x)]    =   0.
x
10
100
1000
10 000
100 000
f(x)
1.080  000
0.091 525
0.009 015
0.000  900
0.000 090
limx→ ∞  [27x/(3x2−5x)]    =   0
Use a graphing calculator/program to graph the function and approximate the limit at infinity.

lim
x → − ∞ 
  6x2+2x

3x2−7x+5
  • The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinity-that is, as x gets very, very large. [Notice that not all functions will have a limit at infinity.] For this problem, we're considering the limit as x → −∞, so we're considering what happens to the function as x goes to very, very large negative numbers.
  • One way to find the value of this limit (or if it even exists) is by using a graphing calculator to graph the function. Using a graph of the function, we can see if the function "settles down" as x→ −∞. Looking at the graph very far to the left, if the function seems to be heading towards a specific value the limit will be that value. On the other hand, if the graph never "settles down" and just keeps growing or changing, then the limit does not exist.
  • Do what the problem tells you to do and graph the function with a graphing calculator/program. You'll get something like the below.
  • Notice that, while interesting, it's not quite as useful as we'd like. We're interested in x→ − ∞, but the left-most value is a paltry x=−15. If we're going to be confident, we should look much farther to the left. Instead, change the Window Settings so that the minimum x-value is much more distant. Something like x=−100 or x=−1000. While we're at it, notice that we don't really care what happens on the right side of the graph, so we can use a much smaller value over there. Finally, don't set your y-values to be too large: from our initial graph, we can see that the limit (if it exists) probably won't get up too high. With this in mind, set the y-minimum at −1 and the y-maximum at 5. Having changed the Window Settings, graph the function again. Now you can use Trace to find values or draw a horizontal line to see what number the function is approaching as it goes far to the left.
limx → − ∞  [(6x2+2x)/(3x2−7x+5)]    =   2
Evaluate the limit (if it exists):    limx → ∞  [3/(x−7)]
  • The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinity-that is, as x gets very, very large. This means that we want to think in these terms: imagine x becoming very large and think about what effect it has on the rest of the function.
  • From the lesson, we learned that if the function we are trying to evaluate is set up in the form of a rational function (polynomial numerator and denominator), we can figure out the infinite limit based on the degrees of the top and bottom. Both the numerator of 3 and the denominator of x−7 are polynomials, so we can do this here. Notice that 3 has a degree of 0 (since it has no variable), while x−7 has a degree of 1 (since it has the variable raised to the first power: x=x1). Thus we have that the numerator's degree is less than the denominator's degree (0 < 1), so the lesson told us that the infinite limit comes out to 0.
  • An alternative way to do this is to think in terms of "What happens to the function as we use BIG NUMBERS for x?" We can symbolize this idea of BIG NUMBERS by using ∞. [Notice that ∞ is not actually a number. It is the idea of having BIG NUMBERS that grow forever, not a value we can actually plug in. Because of this, the below is not formal mathematics. We are playing fast and loose with the math here: while it will make sense to us, this method may not be acceptable in every single situation. Most teachers will accept it, but it is not guaranteed.] With this in mind, we plug in our BIG NUMBER (as ∞) and see what happens:

    lim
    x → ∞ 
      3

    x−7
        ⇒     3

    ∞−7
    Here, notice two things: 1) ∞ minus any constant is unaffected since any number is tiny compared to ∞, so we can throw out the constant; 2) Dividing any constant by ∞ turns to 0: any number is tiny compared to ∞, so it will crush any constant in division.
    3

    ∞−7
        ⇒     3


        ⇒     0
limx → ∞  [3/(x−7)]   =  0
Evaluate the limit (if it exists):    limx → ∞  [(1−5x)/(x+4)]
  • The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinity-that is, as x gets very, very large. This means that we want to think in these terms: imagine x becoming very large and think about what effect it has on the rest of the function.
  • From the lesson, we learned that if the function we are trying to evaluate is set up in the form of a rational function (polynomial numerator and denominator), we can figure out the infinite limit based on the degrees of the top and bottom. Both the numerator of 1−5x and the denominator of x+4 are polynomials, so we can do this here. Notice that 1−5x has a degree of 1 (x=x1), as does x+4. Thus we have that the numerator's degree equals the denominator's degree (1=1). At this point, the lesson said that we need to make a ratio out of the coefficients attached to the variable with the highest exponent:
    1−5x   ⇒  Coefficient: −5                      x+4   ⇒  Coefficient: 1
    Now we make a ratio of dividing the numerator's coefficient by the denominator's coefficient to find the limit:
    −5

    1
        =     −5
  • An alternative way to do this is to think in terms of "What happens to the function as we use BIG NUMBERS for x?" We can symbolize this idea of BIG NUMBERS by using ∞. [Notice that ∞ is not actually a number. It is the idea of having BIG NUMBERS that grow forever, not a value we can actually plug in. Because of this, the below is not formal mathematics. We are playing fast and loose with the math here: while it will make sense to us, this method may not be acceptable in every single situation. Most teachers will accept it, but it is not guaranteed.] With this in mind, we plug in our BIG NUMBER (as ∞) and see what happens:

    lim
    x → ∞ 
      1−5x

    x+4
        ⇒     1−5·∞

    ∞+4
    Now notice that ∞ plus or minus any constant is unaffected since any number is tiny compared to ∞, so we can throw out the constant.
    1−5·∞

    ∞+4
        ⇒     −5 ·∞


    Finally, remember that the ∞ on the top and the ∞ on the bottom represent the same BIG NUMBER so we can cancel them out:
    −5 ·∞


        ⇒     −5
limx → ∞  [(1−5x)/(x+4)]    =   −5
Evaluate the limit (if it exists):    limx → −∞  [(x2)/(100x−70)]
  • The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinity-that is, as x gets very, very large. For this problem, we're working with x→ − ∞ so we want to think in these terms: imagine x becoming a very large, negative number and think about what effect it has on the rest of the function.
  • From the lesson, we learned that if the function we are trying to evaluate is set up in the form of a rational function (polynomial numerator and denominator), we can figure out the infinite limit based on the degrees of the top and bottom. The numerator of x2 and the denominator of 100x−70 are polynomials, so we can do this here. Notice that x2 has a degree of 2 (x2), while 100x−70 has a degree of 1 (x=x1). Thus we have that the numerator's degree is greater than the denominator's degree (2 > 1), so the lesson told us that the limit at infinity does not exist.
  • An alternative way to do this is to think in terms of "What happens to the function as we use BIG NUMBERS for x?" We can symbolize this idea of BIG NUMBERS by using ∞. [Notice that ∞ is not actually a number. It is the idea of having BIG NUMBERS that grow forever, not a value we can actually plug in. Because of this, the below is not formal mathematics. We are playing fast and loose with the math here: while it will make sense to us, this method may not be acceptable in every single situation. Most teachers will accept it, but it is not guaranteed.] With this in mind, we plug in our BIG NUMBER (as ∞) and see what happens:

    lim
    x → −∞ 
      x2

    100x−70
        ⇒     (−∞)2

    100·(−∞)−70
        ⇒     2

    −100·∞− 70
    Now notice that ∞ plus or minus any constant is unaffected since any number is tiny compared to ∞, so we can throw out the constant.
    2

    −100·∞− 70
        ⇒     2

    −100 ·∞
    Furthermore, notice that we can split ∞2 into ∞·∞, then cancel out one of them:
    2

    −100 ·∞
        ⇒     ∞·∞

    −100 ·∞
        ⇒    

    −100
    Finally, dividing ∞ by any constant will have no effect: ∞ is just too big to notice being divided by a normal number. (Still, the negative will have an effect.)

    −100
        ⇒     −

    100
        ⇒     − ∞
    At this point we're done, but we need to interpret what −∞ means. Remember that ∞ is a stand-in for ever growing BIG NUMBERS. Thus, because the result will always have growing BIG NUMBERS, it is never "settling down" to a single value. Instead, as the function goes farther and farther left, it continues to go down forever. This means that the limit does not exist.
limx → −∞  [(x2)/(100x−70)] does not exist
Evaluate the limit (if it exists):    limx → ∞  [(4x3−7)/(x2−2x3)]
  • The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinity-that is, as x gets very, very large. This means that we want to think in these terms: imagine x becoming very large and think about what effect it has on the rest of the function.
  • From the lesson, we learned that if the function we are trying to evaluate is set up in the form of a rational function (polynomial numerator and denominator), we can figure out the infinite limit based on the degrees of the top and bottom. Both the numerator of 4x3−7 and the denominator of x2−2x3 are polynomials, so we can do this here. Notice that 4x3−7 has a degree of 3 (x3), as does x2−2x3. Thus we have that the numerator's degree equals the denominator's degree (3=3). At this point, the lesson said that we need to make a ratio out of the coefficients attached to the variable with the highest exponent:
    4x3−7   ⇒  Coefficient: 4                      x2−2x3   ⇒  Coefficient: −2
    Now we make a ratio of dividing the numerator's coefficient by the denominator's coefficient to find the limit:
    4

    −2
        =     −2
  • An alternative way to do this is to think in terms of "What happens to the function as we use BIG NUMBERS for x?" We can symbolize this idea of BIG NUMBERS by using ∞. [Notice that ∞ is not actually a number. It is the idea of having BIG NUMBERS that grow forever, not a value we can actually plug in. Because of this, the below is not formal mathematics. We are playing fast and loose with the math here: while it will make sense to us, this method may not be acceptable in every single situation. Most teachers will accept it, but it is not guaranteed.] With this in mind, we plug in our BIG NUMBER (as ∞) and see what happens:

    lim
    x → ∞ 
      4x3−7

    x2−2x3
        ⇒     4(∞)3−7

    (∞)2−2(∞)3
    Now notice that ∞ plus or minus any constant is unaffected since any number is tiny compared to ∞, so we can throw out the constant. Along similar lines of thought, but considerably more advanced, notice that ∞2 is massively smaller than ∞3. In a way, because we have a larger "class" of ∞, we can throw out the smaller scale one that is being added/subtracted.
    4·∞3−7

    2−2·∞3
        ⇒     4·∞3

    −2·∞3
    Finally, remember that the ∞ on the top and the ∞ on the bottom represent the same BIG NUMBER so we can cancel them out:
    4·∞3

    −2·∞3
        ⇒     4

    −2
        ⇒     −2
limx → ∞  [(4x3−7)/(x2−2x3)]    =   −2
Evaluate the limit (if it exists):    limx → −∞  cos( [1/x])
  • The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinity-that is, as x gets very, very large. For this problem, we're working with x→ − ∞ so we want to think in these terms: imagine x becoming a very large, negative number and think about what effect it has on the rest of the function.
  • In the lesson, we did not specifically learn how to deal with trigonometric functions like cosine. However, we did learn that

    lim
    x → −∞ 
    1

    x
        =     0
    With this idea in mind, we see that since we have cosine acting on [1/x], when we consider the limit at infinity for cos([1/x]), we can swap out the [1/x] for what it becomes in the long run:

    lim
    x → −∞ 
     cos
    1

    x

        =    
    lim
    x → −∞ 
     cos( 0)
    From there, we see that cos(0) is unaffected by the limit (because it no longer contains the variable x), and we can just evaluate it as normal.

    lim
    x → −∞ 
     cos( 0)     =     cos( 0)     =     1
  • An alternative way to do this is to think in terms of "What happens to the function as we use BIG NUMBERS for x?" We can symbolize this idea of BIG NUMBERS by using ∞. [Notice that ∞ is not actually a number. It is the idea of having BIG NUMBERS that grow forever, not a value we can actually plug in. Because of this, the below is not formal mathematics. We are playing fast and loose with the math here: while it will make sense to us, this method may not be acceptable in every single situation. Most teachers will accept it, but it is not guaranteed.] With this in mind, we plug in our BIG NUMBER (as ∞) and see what happens:

    lim
    x → −∞ 
     cos
    1

    x

        ⇒     cos
    1

    −∞

        ⇒     cos(0)     ⇒     1
limx → −∞  cos( [1/x])    =   1
Evaluate the limit (if it exists):    limx → ∞  [ [3x/(2x−1)] + [(10x2+3)/((2x−3)2)]]
  • The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinity-that is, as x gets very, very large. This means that we want to think in these terms: imagine x becoming very large and think about what effect it has on the rest of the function.
  • Before we try to evaluate the limit, it will make it a little bit easier to expand things so we can better apply the ideas we've learned:

    lim
    x → ∞ 
     
    3x

    2x−1
    + 10x2+3

    (2x−3)2

        =    
    lim
    x → ∞ 
     
    3x

    2x−1
    + 10x2+3

    4x2−12x+9

  • Now that we clearly have two rational function expressions (polynomial numerator and denominator), we can apply what we learned before. Deal with each part of the sum separately:

    lim
    x → ∞ 
      3x

    2x−1
        ⇒    
    Num. degree:
    1
    Denom. degree:
    1
        ⇒    
    N. coefficient:
    3
    D. coefficient:
    2
    Since the numerator and denominator have matching polynomial degrees, we find the value of the limit by taking the ratio of the leading coefficients:

    lim
    x → ∞ 
      3x

    2x−1
        =     3

    2
    Next, the other half of the limit:

    lim
    x → ∞ 
      10x2+3

    4x2−12x+9
        ⇒    
    Num. degree:
    2
    Denom. degree:
    2
        ⇒    
    N. coefficient:
    10
    D. coefficient:
    4
    Since the numerator and denominator have matching polynomial degrees, we find the value of the limit by taking the ratio of the leading coefficients:

    lim
    x → ∞ 
      10x2+3

    4x2−12x+9
        =     10

    4
        =     5

    2
    Finally, since we know the infinite limit of each half on its own, we can just plug those in when we take the limit of them together:

    lim
    x → ∞ 
     
    3x

    2x−1
    + 10x2+3

    4x2−12x+9

        =     3

    2
    + 5

    2
        =     8

    2
        =     4
  • An alternative way to do this is to think in terms of "What happens to the function as we use BIG NUMBERS for x?" We can symbolize this idea of BIG NUMBERS by using ∞. [Notice that ∞ is not actually a number. It is the idea of having BIG NUMBERS that grow forever, not a value we can actually plug in. Because of this, the below is not formal mathematics. We are playing fast and loose with the math here: while it will make sense to us, this method may not be acceptable in every single situation. Most teachers will accept it, but it is not guaranteed.] With this in mind, we plug in our BIG NUMBER (as ∞) and see what happens:

    lim
    x → ∞ 
     
    3x

    2x−1
    + 10x2+3

    4x2−12x+9

        ⇒     3·∞

    2·∞−1
    + 10 ·∞2 + 3

    4 ·∞2 −12 ·∞+ 9
    For each fraction, notice that ∞ plus or minus any constant is unaffected since any number is tiny compared to ∞, so we can throw out the constant. Along similar lines of thought, but considerably more advanced, notice that ∞ is massively smaller than ∞2. In a way, because we have a larger "scale" of ∞, we can throw out the smaller scale one that is being added/subtracted in the second fraction:
    3·∞

    2·∞−1
    + 10 ·∞2 + 3

    4 ·∞2 −12 ·∞+ 9
        ⇒     3 ·∞

    2 ·∞
    + 10 ·∞2

    4 ·∞2
    Now we can cancel out common factors of ∞ in each fraction since they just represent the same BIG NUMBER, then simplify:
    3 ·∞

    2 ·∞
    + 10 ·∞2

    4 ·∞2
        ⇒     3

    2
    + 10

    4
        =     3

    2
    + 5

    2
        =     8

    2
        =     4
limx → ∞  [ [3x/(2x−1)] + [(10x2+3)/((2x−3)2)]]   =   4
Consider the sequence with general term an = [((n+2)!)/(n! ·2n)]. What is the limit of the sequence as n→ ∞?
  • Although this question is based around a sequence, it works almost identically to previous problems about limits at infinity. We want to think about what happens to the sequence terms as n heads off towards infinity-that is, as n gets very, very large. Imagine n becoming very large and think about what effect it will have.
  • In the current form, it's a little hard to tell what's going to happen because we have factorials on the top and bottom. Begin by simplifying the general term by canceling out based on how factorials work (if you don't remember factorials, check out the lesson Permutations and Combinations):
    an = (n+2)!

    n! ·2n
        =     (n+2)(n+1)n!

    n! ·2n
        =     (n+2)(n+1)

    2n
        =     n2 +3n +2

    2n
  • In this form, we can more easily think about how the sequence behaves as n→ ∞. Two approaches we could use are plugging in large numbers or graphing the expression: either one of these methods will help us figure out what happens in the long run as n gets very, very large. In previous problems, we've relied on working with rational functions (polynomial numerator and denominator) to figure out infinite limits, but we can't do that here. Since the denominator is 2n, we don't have a polynomial on the bottom, so we can't use the exact same analysis.
  • However, we can still think in terms of "What happens to the expression as we use BIG NUMBERS for n?" We can symbolize this idea of BIG NUMBERS by using ∞. [Notice that ∞ is not actually a number. It is the idea of having BIG NUMBERS that grow forever, not a value we can actually plug in. Because of this, the below is not formal mathematics. We are playing fast and loose with the math here: while it will make sense to us, this method may not be acceptable in every single situation. Most teachers will accept it, but it is not guaranteed.] With this in mind, we plug in our BIG NUMBER (as ∞) and see what happens:

    lim
    n → ∞ 
      n2 +3n +2

    2n
        ⇒     2 + 3·∞+ 2

    2
    Begin by noticing that, for the numerator, the biggest "number" by far is ∞2. Because it is ∞·∞, it is much bigger than 3·∞ and 2, so we can just throw those away without affecting anything:
    2 + 3·∞+ 2

    2
        ⇒     2

    2
    This now brings us to the challenging question: which is bigger, ∞2 or 2? Well, remember, ∞ isn't actually a number: it's the idea of plugging in BIG NUMBERS. With this in mind, we realize that we really want to know which grows faster, n2 or 2n? Here we might think back to the lesson on Exponential Functions and remember that 2n grows extremely quickly (remember the story with the grains of rice on the chessboard from that lesson). While n2 gets big, it is completely outclassed by 2n, and will be "crushed" into nothing. Even if we don't remember our work with exponential functions, we can still think about what happens as the numbers get big. Consider if we used a fairly small "big" number, like n=100: while 1002 is pretty large, it is absolutely tiny in comparison to 2100 (check it with a calculator if you're not sure). This disparity in size will only continue to widen as n gets larger, so we see that the denominator will "crush" the numerator into nothing in the long run. Therefore, we have
    2

    2
        ⇒     0
limn → ∞  [((n+2)!)/(n! ·2n)]   =   0
A publishing house is getting ready to print a book. In order to do so, they must first prepare the book (edit, design, etc.), which will cost a total of $12 000. Once the book is prepared, it will cost the publishing house $1.70 per book they print. (a) What is the average cost per book if they print 10  000 books? What if they print 100  000 books? (b) Give a formula that determines the average cost per book of printing n books. (c) Determine the limit of the average cost per book as n → ∞. Interpret what this means in terms of the problem.
  • In general, the average is determined as
    Average     =     Total amount of "stuff"

    Number of "objects"
    For this problem, the "stuff" is the cost of printing all the books, while the number of "objects" is how many books are printed.
    Average cost per book     =     Total cost of making books

    Number of books made
  • (a): Calculate the cost of producing 10 000 books. Remember, there is an initial, flat cost of $12 000, then an additional $1.70 per book. This means:
    Cost of producing 10 000 books     =     12 000 + (10 000)(1.70)     =     29 000
    Once we know the cost of producing 10 000 books, we want to know the average cost per book, so we divide the total cost by the number of books:
    Average cost per book, printing 10 000     =     29 000

    10 000
        =     2.90
    Thus when we print 10 000 books, the average cost is $2.90 per book.

    We follow the same method to do it for 100  000 books:
    Cost of producing 100 000 books     =     12 000 + (100 000)(1.70)     =     182 000

    Average cost per book, printing 100 000     =     182 000

    100 000
        =     1.82
    Thus when we print 100 000 books, the average cost is $1.82 per book.
  • (b): To find a general formula that determines the cost per book of printing n books, notice that we are determining our average cost per book from the idea
    Average cost per book     =     Total cost of making books

    Number of books made
    Thus, to figure out the formula, we just need to calculate the 1) Cost of making n books; and 2) Number of books made. Well, as we saw in the above step, there is an initial, flat cost of $12 000, then an additional $1.70 per book. Since we are making n books, we have
    Cost of making n books     =     12 000 + 1.70·n
    It's much easier to figure out the number of books made: we're making n books, so that's the number made. Put these two things together to find the average cost per book:
    Average cost per book, printing n     =     12 000 +1.70 ·n

    n
  • (c): Since we now have a formula for the average cost per book, we just need to consider the limit as n→ ∞:

    lim
    n → ∞ 
      12 000 +1.70 ·n

    n
    From the previous problems we did in this lesson, we saw that this will just come down to a ratio of the coefficients on the variables:

    lim
    n → ∞ 
      12 000 +1.70 ·n

    n
        =     1.70

    1
        =     1.70
    Thus, as n→ ∞, the average cost per book goes to $1.70.

    Interpreting this, remember what this all means. The limit as n→ ∞ is a way to ask how much the average cost per book is as we produce HUGE numbers of books. Thinking about it like that, we realize that as the number of books printed becomes extremely large, the initial start-up cost becomes a tiny fraction of the overall cost. Once we're printing on the scale of billions of books, that initial cost to prepare the book becomes negligible in comparison. With huge numbers of books, we no longer need to pay attention to the start-up cost, so it's just a matter of how much it costs to print additional books. Since printing another book costs $1.70, that is what the average cost per book becomes when printing in massive quantities.
(a) $2.90,     $1.82 (b) [(12 000 +1.70 ·n)/n] (c) $1.70-When printing very large numbers of books, the start-up cost becomes negligible, so the only important thing is the cost of producing each book.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Limits at Infinity & Limits of Sequences

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Introduction 0:06
  • Definition: Limit of a Function at Infinity 1:44
    • A Limit at Infinity Works Very Similarly to How a Normal Limit Works
  • Evaluating Limits at Infinity 4:08
    • Rational Functions
    • Examples
    • For a Rational Function, the Question Boils Down to Comparing the Long Term Growth Rates of the Numerator and Denominator
    • There are Three Possibilities
    • Evaluating Limits at Infinity, cont.
    • Does the Function Grow Without Bound? Will It 'Settle Down' Over Time?
    • Two Good Ways to Think About This
  • Limit of a Sequence 12:20
    • What Value Does the Sequence Tend to Do in the Long-Run?
    • The Limit of a Sequence is Very Similar to the Limit of a Function at Infinity
  • Numerical Evaluation 14:16
    • Numerically: Plug in Numbers and See What Comes Out
  • Example 1 16:42
  • Example 2 21:00
  • Example 3 22:08
  • Example 4 26:14
  • Example 5 28:10
  • Example 6 31:06

Transcription: Limits at Infinity & Limits of Sequences

Hi--welcome back to Educator.com.0000

Today, we are going to talk about limits at infinity and limits of sequences.0002

Up until this point, we have only considered the idea of a limit as x goes to c, where c is some fixed horizontal location.0006

But what if, instead of focusing on a single location, we considered what would happen to the function if x just kept traveling forever off to infinity?0013

What would we do if our x wasn't going to a single place, but we were just sort of watching it ride off into the sunset?0020

This is the idea that we will consider in this lesson.0026

In fact, we have actually considered this many lessons ago, in the lesson on horizontal asymptotes,0029

because a horizontal asymptote was the question of what this function goes to in the very, very long run;0034

as x runs off to infinity (both the positive and the negative infinity)0040

what vertical value, what y-value, what function value, do we end up running to in the long run?0046

That was the idea of a horizontal asymptote; we will draw on those ideas in this lesson, as well.0052

Before we start, though, let's remind ourselves of something important: infinity is not a location.0056

It is not some place; you don't run to the end of the rainbow, and there you are at infinity; you always have to keep running.0061

The idea of infinity is just going on forever, moving on to ever-larger numbers.0068

You never actually reach infinity; you can travel towards infinity, but you can never reach infinity.0074

It is just the idea of going on forever; it is the idea of riding into the sunset.0081

You don't ever actually make it to the sun; you just keep riding off into the distance.0085

That is the idea of what it means to travel towards infinity.0090

You never actually arrive there; x can't be equal to infinity; but we can consider the idea of what happens as x runs off forever and ever and ever.0093

That is what we will be thinking about.0101

We denote the limit of a function at infinity (even though it says "at," we are really meaning "as it goes towards")--0104

the limit as x goes to negative infinity of f(x) and the limit as x goes to positive infinity of f(x)...0111

And this means the value that f(x) approaches as x goes off to either negative infinity or positive infinity, respectively.0118

So, negative infinity, with an actual negative sign--that means negative infinity.0127

And infinity, just on its own with nothing there--we just assume that there is a positive sign in front of it, even though we don't see it.0133

If you don't see a symbol, it is assumed that we are talking about positive infinity, as opposed to talking about negative infinity.0140

Negative infinity would go off to the left, whereas positive infinity, or just infinity with no symbol in it, goes off to the right, forever.0146

OK, a limit at infinity works very similarly to how a normal limit works.0154

Does the function settle down--does it go to some specific value l?0160

It is just that, in this case, we are talking about long-term behavior, instead of x going to some specific horizontal location.0166

Instead of what happens to the function as x gets close to c, it is what happens to the function as it rides off into the sunset.0173

What happens as it goes off to some infinity?0179

Now, it is important to note that most of the functions--in fact, the vast majority of the functions0181

that we are used to working with--do not have limits at infinity.0186

For example, if we consider f(x) = x, one of the simplest functions that we are used to using,0189

that one just keeps growing forever; so it has no limit at infinity.0194

It doesn't stabilize; it doesn't settle down to some value.0199

If you plug in 1 million, you get 1 million out of it; if you plug in 1 billion, you get 1 billion out of it; if you plug in 1 trillion, you get 1 trillion out of it.0204

As you keep plugging in larger and larger numbers, it is just going to keep growing and growing and growing and growing.0211

It is never going to stop growing, which means it is never going to settle down, which means it is never going to some specific value l.0215

Most of the functions that we are used to dealing with on a daily basis are actually not going to have limits at infinity,0222

because they never settle down to a specific value--they grow without bound.0227

They might grow off to positive infinity; they might grow off to negative infinity (that is, vertically--what the output goes off to be).0231

But they are growing without bound; they aren't going to some specific value, so that means that they won't have limits at infinity.0237

Still, there are definitely functions that do have limits at infinity.0243

The type of functions that we will work with most often (there are some others that won't be this, but)0247

the ones that we will work with most often, that have limits at infinity, are rational functions.0252

We worked with these many lessons ago, when we learned about asymptotes.0256

They are functions of the form f(x) = n(x)/d(x), where n(x) and d(x) are polynomials, and d(x) is not equal to 0.0260

So, you probably remember these things like this: g(x) = (3x - 1)/(x3 + 4), h(x) = 1/x4,0271

j(x) = (x5 + 47x2)/(x3 - 15)--just some polynomial divided by some other polynomial.0280

Now, because we are dividing by something, that means that our denominator, what we are dividing by,0289

has the possibility to grow faster than the numerator.0294

Basically, our denominator can grow fast enough to keep the numerator in check--to keep that numerator from blowing off and just going on forever.0297

The denominator can actually grow faster and keep it down.0305

It has the ability to stabilize it to a single value in the long run.0309

And that is why we end up seeing rational functions give us limits at infinity so often.0312

For a rational function, the question is basically comparing the long-term growth rates of the numerator and the denominator.0317

It is a question of which is growing faster over the long term: is it the numerator or the denominator?0323

If the numerator is growing faster than the denominator over the long term, then the thing is not really going to settle down,0329

because it is just going to keep getting bigger and bigger and bigger.0335

If, on the other hand, the denominator is growing faster than the numerator,0338

then the denominator will crush the numerator, and so it will be forced to settle down.0341

Now, we already studied this idea in horizontal asymptotes; and so, let's look at those results.0346

If we have some rational function, f(x) (and notice that that is just some polynomial divided by some polynomial--0352

some constant times xn, some other constant times xn - 1, and working our way0358

down to a constant times x, plus some constant, and the same thing on the bottom, as well--0364

it is just constant times x to some value, constant times x to that value minus 1,0369

and working our way down to a constant; so it is some polynomial over some polynomial),0374

notice, from this, that n is the numerator's degree; we have xn as the largest exponent on the top;0378

and m is the denominator's degree, so m is the biggest exponent on the bottom.0387

From this, there are three possibilities: if n is less than m, then that means that our top,0394

the exponent in our numerator, n, is going to be less than the exponent in our denominator,0401

which means that the denominator is going to grow faster, so it will crush the numerator, causing us to have a horizontal asymptote of y = 0.0408

If, on the other hand, n equals m (the leading exponent on the numerator is equal0415

to the leading exponent on the denominator), they grow in the same category of speed.0423

They won't necessarily have precisely the same; but one of them won't massively outclass the other one.0427

At that point, what we will do is compare the leading coefficients, an and bm.0432

The horizontal asymptote that we get out of that is a ratio of the leading coefficients, an divided by bm,0440

because in the long run, since we have the same exponent on top and bottom,0447

that part, the x to the some exponent, will grow at the same rate on the top and the bottom.0451

So, it will end up just being a question of what number they are multiplying in front.0455

And that is why we get a horizontal asymptote based on that.0458

And finally, the last one: if n is greater than n (that is, the leading exponent on the numerator is greater0461

than the leading exponent on our denominator), then that means that the numerator will be able to run faster0467

than the denominator and escape the denominator's ability to bound it and hold it back.0471

And so, it will just go off forever, and it won't be able to stabilize to a single value, which means that it will have no horizontal asymptote.0476

We can write this in a way where we can talk about this as limits at infinity, because,0484

since horizontal asymptotes tell us the behavior of f(x) as x goes to positive or negative infinity--0489

a horizontal asymptote is what value it approaches over the long term--they are also telling us the limits at infinity,0496

since the limit at infinity is what value it approaches over the long term.0501

So, for some rational function f(x), let n be the numerator's degree, and m be the denominator's degree.0505

Let an and bm be the leading coefficients of the numerator and the denominator, respectively.0514

Then, we have: if n is less than m, the limit as x goes to positive or negative infinity of f(x) equals 0.0524

The numerator is smaller, effectively; the exponent is smaller, so its growth rate is smaller than the denominator.0532

So, the denominator crushes it down to 0.0538

If n equals m, then the limit as x goes to positive or negative infinity of f(x) is equal to an divided by bm.0541

The growth rate on the top and the bottom is the same, because they have the same leading exponent.0547

So now, it is a question of what the ratio is of the coefficients in front of them.0551

And finally, if n is greater than m, that means that the leading coefficient on top is greater than the leading coefficient on the bottom,0555

which means that the growth rate of the top is greater than the growth rate of the bottom.0560

So, there the top manages to escape and run off forever.0564

So, that means that the limit as x goes to positive or negative infinity of f(x) simply does not exist, because it will never stabilize to a single value.0568

That tells us what to do with rational functions; the previous method allows us to easily find limits at infinity for rational functions.0577

But we will occasionally have to deal with other types of functions, as well.0583

We won't only have to deal with rational functions.0586

So, in that case, the best thing to do--there is no simple formulaic method for how to figure out "Here is what the limit is going to end up being."0589

In this case, what you want to do is think in terms of how the function will be affected as x grows very large, both positively and negatively.0596

You want to think, "Does the function grow without bound?"0605

If it just grows forever and ever, or goes off down forever and ever,0608

then that means that it is going to end up not stabilizing to something in the long term, which means that it won't have a limit at infinity.0612

Or, on the other hand, will it settle down--does it go to some specific value?0617

Does it settle down; does it approach some specific thing over the long term?0621

So, there are two good ways to think about this, to figure out if this is the case.0625

We want to think about what happens if we plug in a large number.0628

And by "large number," I mean to think like...if I were to plug in something on the scale of a million,0632

a billion, a trillion, a really, really, really big number--something large--what would happen to this?0637

You don't have to come up with actual answers to what will happen to the thing.0643

You don't have to produce some number in the end.0647

You just want to think, "If there was a really, really big number here, how would it affect the other things that it is near?"0649

What would no longer really be important? What would still matter?0654

If you were to plug in a really big number, what is going to keep changing?0658

And what will happen as that large number continues to increase?0660

That is one way of looking at it; the other way to look at it is to think, "What are the rates of growth in the function?"0664

Which part of the function is growing faster and will continue to grow?0670

Which parts grow faster, and which parts are growing slower, that get slower and slower as we go farther and farther out?0674

Thinking about these two things (that one will especially help if you are dealing with a fraction--0681

what is the comparison between the growth rate of our numerator versus the growth rate of our denominator?)--0686

thinking in terms of rates of growth, what is growing faster and what is growing slower,0690

how the rate of growth will be affected as we go out to larger and larger x--0694

these sorts of things (that and what would happen if I plugged in a very large number)--0698

thinking in terms of those two ideas will give you a good, intuitive sense of what is going to happen.0702

If you think through these questions, you can get a good idea of where the function will be going,0707

of how the function will behave over the long term.0711

You won't necessarily be able to come up with an absolutely precise answer.0715

But you will be able to get a sense of if it makes sense for this thing to have a limit at all.0717

But sometimes, you will even be able to get an exact answer by thinking through this; it depends on the situation.0721

But just sort of try to be creative and think in a very broad, general sense.0726

We will also talk about ways where, if you are not quite sure how to think about it,0730

there are numerical ways that you can figure out...get a good sense of what is going on.0733

And we will talk about that in just a few slides.0737

Another thing that we can talk about, using this idea of a limit at infinity, as a limit goes to infinity: we can apply it to a sequence, as well.0740

If we have some sequence, a1, a2, a3, a4, a5...0747

so it is some infinite sequence that just keeps going on forever and ever and ever--0752

then what we can consider is the limit as n goes to infinity of an.0756

What does our sequence go towards--what value does the sequence approach in the long run?0760

How does this thing work out? What will it be going towards as n becomes ever larger and larger and larger?0767

The limit of a sequence (that is this thing right here) is very similar to the limit of a function at infinity.0772

The question is, "Does the sequence settle down--does it go to some specific value l as n runs off to infinity?"0779

As our n becomes larger and larger and larger, does our sequence stabilize0787

into something that is going to basically be the same as we go farther and farther and farther?0790

Now, it is important to note that, just like functions, most sequences will not have limits as n goes to infinity.0795

For example, a really simple sequence: 1, 2, 3, 4, 5, 6... has no limit, because it just grows forever.0801

It will just grow forever, because we have 1, 2, 3, 4...so if we look at a very far-out term, it will be very large.0810

But if we look at an even farther-out term, it will have continued to grow, and it will be even larger.0817

So, it is not going to head towards a steady value; it is not going to stabilize and go to some specific value l; it will never settle down.0821

Nonetheless, there are definitely still some sequences out there that will end up stabilizing; and we will see those in the examples.0830

But just because we are looking at the limit as n goes to infinity of a sequence doesn't necessarily mean that it will stabilize.0836

There are plenty of sequences out there that won't stabilize at all.0842

For example, every arithmetic sequence we have ever looked at won't stabilize,0845

because it just continues stepping up and stepping up and stepping up and stepping up.0849

Finally, we can also talk about numerical evaluation.0855

Sometimes, it can be difficult to tell how a function or sequence will behave in the long run.0858

In that case, we can evaluate the function numerically--that is to say, use numbers.0862

We will just plug in numbers, and we will see what comes out.0866

If we have a calculator, we can use a calculator and just plug in very large numbers.0869

And we will want to plug in both positive and negative numbers.0874

And then, see what happens: we just see what happens to our function or a sequence.0876

We plug in 10; then we plug in 100; then we plug in 1000; then we plug in 10000.0880

Does it seem like it is going to a number, or does it seem like it is just growing larger and larger and larger?0884

Doing this will give us a good sense for long-term behavior.0889

We will be able to tell that, yes, it seems to be just growing forever and ever,0893

or it seems to be stabilizing as we go to larger and larger numbers that we are plugging in.0896

So, this gives us a way to numerically get a sense of what is going on.0900

It is not foolproof, but for the most part you will be able to figure out which one it is going to end up going to.0903

And you probably also have a very good estimate of what value it is going to be approaching in the long term.0908

Similarly, if you have access to a graphing calculator or some graphing program, you can graph the function.0913

If you expand the viewing window to a large horizontal region (say -100 to +100),0920

then you can look and see if the graph settles down in the long run.0925

Does it seem like it is being pulled to a single value, or does it seem like it is just blowing forever and going to keep growing forever and ever and ever?0929

Now, once again, it is not a foolproof method; there are some times where the function will fool you for the first 1000x.0935

From 0 to 1000, it will look like it is growing forever and ever.0941

But then, after 1000, it will actually end up steadying out to a single value.0944

But for the most part, this is a pretty good way to see if this is going to end up approaching a single value,0948

or if it is going to just keep growing forever and ever.0953

So, just take a look at the graph; make sure you use a large horizontal region.0955

If you only look at -10 to +10 for x, you might not have a very good sense of what happens in the long run.0960

You want to use a very large horizontal region, like -100 to +100.0965

It might be kind of hard for your graphing calculator to graph as you get to larger and larger windows;0969

but the larger you can deal with, the better, really, because that will tell you a better idea of what is going on.0973

For the most part, though, -100 to 100 should probably do for anything that you want to graph.0978

And then, just look: does the graph settle down--it is going to tend to a single value in the long run?0982

As you go to those larger x-values, is it basically always graphing at the same height?0987

And if that is the case, it probably has a limit; and you can figure out, looking at the graph, about what value that ends up being.0991

All right, we are ready for some examples.0998

The first example: Evaluate the limits below if they exist.1000

The first one is the limit, as x goes to negative infinity, of 1/x.1003

In this case, we want to think about what ends up happening.1007

We have that specific formulaic method; we have a step-by-step thing for analyzing what is going to come out here.1010

We already can see that the answer has to be 0, from that formulaic method.1015

But let's also think about what happens: as x goes off to negative infinity, our 1/x...1018

well, x continues to grow; how does the numerator grow?1025

It doesn't grow; it is just 1--it stays constant; it just sits there as 1 forever and ever and ever.1029

But our x continues to get larger and larger and larger.1033

We plug in 1000 (well, we plug in -1000, because we are going to negative infinity), and we have 1/-1000;1036

we plug in negative 10 thousand, and we have 1 over negative 10 thousand;1043

we plug in negative 1 billion, and we have 1 over negative 1 billion.1045

We are getting smaller and smaller and smaller; it is crushing down to 0, so we see that it ends up being 0.1050

We can also tell that that is going to end up being the case, just because our numerator has a leading exponent of 0.1056

It is x0, which would only give us 1; and our bottom has x1, so the bottom has a higher leading exponent;1062

so that means that it is going to be crushed down to 0 in the long run.1068

Similarly, the limit as x goes to positive infinity of 1/x...well, our bottom is x; it does grow;1072

but our top is 1--it just stays the same forever and ever.1079

So, that means, as x gets larger and larger and larger, that it is going to crush our entire fraction down to 0.1081

So, we end up getting 0 for the limit here.1087

Over here, we have the limit as x goes to negative infinity of (x3 + 3)/(x2 + x).1091

So, notice: in this case, we have 3 as the leading exponent up top.1097

That means that we have this growth rate somewhere in the neighborhood of x3.1102

But on the bottom, we have a leading exponent of squared; so we have a growth rate somewhere in the neighborhood of x2.1106

What that means is that the top, in the long run, will end up growing much, much, much faster than our bottom will.1112

It is going to end up outrunning the bottom, effectively.1119

We can also imagine this: if we were to plug in a very large number,1122

then we would have big cubed, plus 3, over big squared, plus big.1125

Well, notice: the number here that is the most important is big cubed.1135

Big squared is a very large number, but big cubed is going to be even larger.1141

There is a huge difference between 102 and 103; 10 squared is 100, but 10 cubed is 1000.1147

So, as we get out to very large numbers, big cubed is going to be massively larger than big squared.1152

And similarly, big is just to the 1; so it is practically not going to be anything compared to big squared.1157

So, in the long run, the plus 3 doesn't really matter; the big to the 1 doesn't really matter.1162

The big squared doesn't really matter, because the biggest thing of all, by far, is big cubed.1167

So, that means that we are going to get really, really large numbers up top,1172

and nothing else is really going to be of a comparative size.1175

So, that means, in the long run, that it is going to blow out to infinity.1178

In this case, it will blow out to negative infinity; so that means that it is not stabilizing to a single value.1181

So, we say that the limit does not exist, because it is never going to stabilize,1186

because our top grows faster than our bottom will.1190

In this case, it is going to negative infinity, so we might think of it as growing down.1195

But in either case, it is going larger than the bottom will.1199

The final one is the limit as x goes to positive infinity of (8x4 + 3x2)/(2x4 - 17).1202

So, in this case, we see that the important thing is a leading coefficient of 4 on top, and a leading coefficient of 4 on the bottom.1210

3x2 and -17...as we get to very large numbers, as we go farther and farther out towards infinity,1215

3x2 and -17...they aren't really going to matter much in the long run, as we get to very large numbers.1222

So, it really is determined by 8x4/2x4.1227

In that case, we see that the x4 and the x4 are going to effectively cancel each other out.1231

So, all that we really have left, in the long run, is 8/2; 8/2 simplifies to 4, and there is our answer.1235

We can also see this as the leading coefficients, 8 and 2; since we have the leading exponents,1242

we just do the leading coefficient on the top, divided by the leading coefficient on the bottom, 8/2; and that is equal to 4; great.1248

The next example: Let's look at the limits here--these are limits of sequences, since it is n going to infinity.1256

Evaluate the limits below, if they exist: limit as n goes to infinity of 1/n2.1262

Once again, we see that this is n2; up at the top, it is just 1; it is constant.1267

So, our bottom continues to grow and grow and grow and grow, but our top stays the same.1272

So, as we divide by larger and larger and larger numbers, it crushes it down to 0, just like the reasoning that we used previously.1276

The limit as n goes to infinity of (5n - 1)/(n + 4): well, in this case, we have 5n and n over here.1282

The -1 and the 4 don't change as n goes larger and larger and larger.1291

In the long run, we have big numbers for n; -1 and 4 are going to basically have no effect on what is going on.1296

So, we can think of them as not really mattering, which leaves us with 5n/n in the long run;1302

so we are just comparing--what are the two leading coefficients?1307

5/1 = 5; and there is the limit as n goes out to infinity, the limit of the sequence--what happens to the sequence in the long run.1311

Compare the limits below: which limit exists? Why?1323

All right, our first one is the limit as x goes to infinity of sin(x), and our second one is the limit as x goes to infinity of sin(x)/x.1326

OK, let's get a sense for what happens to the limit as x goes to infinity of sin(x).1334

Well, first let's take a quick graph of how sin(x) behaves.1338

We start here at x = 0; it goes up and down and up and down and up and down and up...1342

and it just continues in this method forever and ever and ever and ever.1354

It never changes this thing of going up/down, up/down, up/down; that is how sin(x) works--it repeats itself over and over forever.1358

What that means is that we have it bouncing; we are bouncing between +1 at its maximum and -1, forever.1366

We are always going up/down, up/down, up/down, up/down; we never stop bouncing up and down.1377

So, if that is the case, since we never stop bouncing up and down, it never settles down to a specific value.1384

All right, it is going to always be near the values of +1 and -1 and 0; but it never steadies out to a single thing.1391

If we say that it is going to be at 0 in the long run, well, it is going to end up getting away from 0 over and over and over.1399

So, it is never settling down; if it never settles down, that means that the limit does not exist; the limit here does not exist.1405

What about our other limit, though--the limit as x goes to infinity of sin(x)/x?1417

Well, what happens? Once again, sin(x), our top, bounces between -1 and +1 forever.1423

OK, but the bottom grows forever; this x right here is going to get larger and larger and larger as x goes off to infinity.1437

So, as x goes off to infinity, the bottom will grow forever.1450

The top oscillates between +1 and -1, +1 and -1, +1 and -1; but our bottom gets larger and larger and larger: 1, 10, 100, 1000...1455

So, since the top never really manages to get very far--it isn't growing without bound--1465

it is just bouncing between these two numbers--even at its largest possible values of +1 and -1,1470

if we divide that by x out at a billion, x out at a quadrillion...it is going to be crushing it down to these very small numbers.1475

Thus, we have that the fraction will end up being crushed.1483

The bottom, in the long run, is going to crush the top.1487

In the long run, it ends up looking like 0.1497

If you want to see what that ends up looking like, what we have is this divide by x...well, 1/x has a graph like this, as it approaches it.1500

So, our sin(x) is going to be bouncing between these two possible extremes.1515

So, it ends up getting squeezed down, closer and closer and closer and closer to this 0 value.1520

And that is why it ends up happening--that is why you end up having this long-term value of 0.1526

In sin(x), it bounces up and down forever; it just keeps going up and down and up and down.1532

But over here in sin(x) divided by x, this "divide by x," over the long run, pinches it down--keeps it crushed down.1537

It starts with these large oscillations; but as it goes farther and farther out, it has to get smaller and smaller,1544

because the x, the "divide by x," crushes it down; and so it gets crushed down to a single value.1548

It will continue to oscillate, but it is getting closer and closer and closer; it has to stay in this window near this height value of 0.1553

And so, since it gets crushed down slowly over time to 0, it effectively just approaches 0 in the long run.1560

So, we have a limit as x goes to infinity of 0.1566

The fourth example: Compare the limits below. Which exists? Why?1569

First, we could just graph this to get a sense of what is going on.1574

If we graph this, the limit as x goes to negative infinity of 2x, and the limit as x goes to positive infinity of 2x...1577

well, if we graph, what does 2x look like?1584

Well, at 0, it is going to be at 1; and then as we go out, it is going to get very large very quickly.1586

As we go to the left, it is going to get smaller and smaller and smaller and smaller.1590

All right, that is what happens: it will never get past the x-axis, but it is going to end up getting smaller and smaller and smaller.1594

If we look at some values, we see that at x = -1, x = -2, x = -3...for this, we would have 2-1,1600

and then 2-2, and then 2-3, which would come out to be 1/2, 1/22, so 1/4,1609

1/23, which would be 1/8; so 1/2, 1/4, 1/8...it gets smaller and smaller.1616

It is always going to smaller values as x goes off to negative infinity.1624

Since it is always going off to smaller and smaller values--that is, values closer and closer to 0--1628

as 2 to the negative number becomes very large--it is going to be1633

1 over 2 to the very large number, which is going to make a very tiny fraction.1638

So, over the long run, it ends up getting crushed down to 0.1642

However, if we look at the limit as x goes to positive infinity of 2x,1646

if we look at just the first couple of numbers, 21, and then 22,1650

and then 23 (that is, x = 1, x = 2, x = 3), we would end up getting 2, and then 4, and then 8;1655

so it is getting bigger and bigger as it ends up going larger and larger.1661

As it gets closer and closer to positive infinity, it will get larger and larger and larger.1667

We end up seeing that, since it is going to get larger and larger and larger, it is never going to stabilize to a single value.1671

It is never going to go to some specific value l, so that means that the limit does not exist,1677

because it will just blow off forever and ever, going up forever and ever.1682

The fifth example: Evaluate the limit as x goes to infinity of 2x/(x + 1) - x2/4(x + 1)2.1686

The first thing to notice here is that this portion of this fraction here doesn't really have an effect on this fraction here during the process of the limit.1696

So, as x goes to infinity, this fraction and this fraction don't really interact with each other.1705

They are basically separate; so if they are basically separate, we can split the limit into the two portions.1711

So, we can split it into the limit as x goes to infinity of the first portion, 2x/(x + 1),1716

minus the limit as x goes to infinity of the second portion, x2/4(x + 1)2.1725

All right, now we can evaluate both of these on their own.1736

For the first one, 2x/x...they both have the same leading coefficient.1738

If we imagine very large numbers going in there, we are comparing two times big number, over big number plus 1.1743

The plus one doesn't really matter; so we only care about two times big over big.1748

The "big"s cancel each other out effectively, and we can think of this as just going...it will go to precisely 2 in the long run.1752

As x goes off forever and ever, it is going to get closer and closer to 2.1760

Minus the limit as x goes to infinity...for this one, we are not quite sure, because...1765

let's expand the (x + 1)2, although we can see x21768

divided by something that is also going to contain an x2.1772

So, we should probably be able to see that in the long run, as x goes to positive infinity, we will end up seeing it go to 1/4.1775

But let's expand it, so we can see it clearly: the limit as x goes to infinity...1786

x2 doesn't change on top...divided by 4, times (x + 1)2...that is just equal to x2 + 2x + 1.1789

So, if we multiply 4 times that expansion of (x + 1)2, we get 4x2 + 8x + 4.1800

So, we still have 2 in front, minus the limit, as x goes to infinity, of this.1809

Well, actually, at this point, we don't even need to do another limit,1814

because we can see that the top has a leading exponent of x2; the bottom has a leading exponent of squared, as well.1817

So, we just compare the coefficients in front, 1 and 4.1824

Since we have big number squared up top, divided by 4 times big number squared, plus 8 times big number...1828

that is not really going to be much, compared to big squared...plus just plain 4 (that is not going to be much compared to big),1834

it is really 1 big squared, over 4 big squared; the big squareds effectively cancel out, leaving us with 1/4 in the long run.1839

We have minus 1/4; we have broken down each piece of the limit; we have figured out1847

that the first portion becomes 2; the second portion becomes 1/4; so 2 - 1/4 simplifies to 7/4.1852

All right, the final example, Example 6: Evaluate the limit as n goes to infinity of the sequence (n - 1)!/(n + 1)!.1862

The first thing we want to do here is think, "Well, we don't really see how to do this immediately;1871

so we want to see if we can simplify this into something where we have less going on."1875

Factorials--it is kind of hard to see what is going on with factorials.1879

So, maybe let's get a sense for if there is some way to cancel them and expand things.1882

We realize that they are both based around a somewhat similar thing.1886

n + 1 isn't very far from n - 1; so we can expand the factorials, so that we can cancel out based on that.1891

We have (n - 1)! on top, and (n + 1)!...well, that is going to be n + 1 times one less than that, which is going to be n,1897

times one less than that, n - 1, times one less than that...1906

Well, if we keep going down forever, that is going to be (n - 1)! here.1909

So, we have (n - 1)! on the top and n + 1 times n times (n - 1)! on the bottom.1912

Well, we can cancel the (n - 1)!'s now; and we have the limit as n goes to infinity of 1/(n + 1)(n).1918

So now, we can see that, as n goes off to infinity, well, our top doesn't change at all; it is just a constant in this case.1929

So, since our top isn't ever going to change, but our bottom, (n + 1) times n, is going to get larger and larger and larger1935

as n goes off to infinity, that means our bottom is growing, but our top is just staying the same.1940

So, in the long run, it is going to get crushed down to 0; the fraction will get crushed down to 0, so the limit of this sequence is 0.1944

All right, that finishes our exploration of limits in this course.1951

We are now going to move on to derivatives, and we will get a cool sense for how derivatives work.1955

It is really great stuff; we are getting a chance to see a preview of calculus, which is going to be really useful1958

for when we get to calculus, because we are setting a groundwork here1962

that you will then be able draw upon later, when you learn this stuff again.1964

All right, we will see you at Educator.com later--goodbye!1967