For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
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).  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 longterm 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) = a_{n} x^{n} + a_{n−1} x^{n−1} + …+ a_{1} x + a_{0}
.  If n < m ⇒ lim_{x→ ±∞} f(x) = 0
 If n=m ⇒ lim_{x→ ±∞} f(x) = [(a_{n})/(b_{m})]
 If n > m ⇒ lim_{x→ ±∞} 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?
 We can take this idea of a limit going to infinity and apply it to a sequence as well. Let a_{1}, a_{2}, a_{3},... be some infinite sequence. Then we can consider
lim
a_{n}.  Sometimes it can be difficult to tell how a function or sequence will behave in the longrun. 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 longterm 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 longrun.
Limits at Infinity & Limits of Sequences

 The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinitythat 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 xvalues 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 xvalues, the completed table is
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 lim_{x→ ∞} [27x/(3x^{2}−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





 







 The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinitythat 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 leftmost 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 xvalue 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 yvalues 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 yminimum at −1 and the ymaximum 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.
 The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinitythat 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=x^{1}). 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:
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.
lim
x → ∞3 x−7⇒ 3 ∞−73 ∞−7⇒ 3 ∞⇒ 0
 The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinitythat 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=x^{1}), 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:
Now we make a ratio of dividing the numerator's coefficient by the denominator's coefficient to find the limit:1−5x ⇒ Coefficient: −5 x+4 ⇒ Coefficient: 1 −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:
Now notice that ∞ plus or minus any constant is unaffected since any number is tiny compared to ∞, so we can throw out the constant.
lim
x → ∞1−5x x+4⇒ 1−5·∞ ∞+4
Finally, remember that the ∞ on the top and the ∞ on the bottom represent the same BIG NUMBER so we can cancel them out:1−5·∞ ∞+4⇒ −5 ·∞ ∞−5 ·∞ ∞⇒ −5
 The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinitythat 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 x^{2} and the denominator of 100x−70 are polynomials, so we can do this here. Notice that x^{2} has a degree of 2 (x^{2}), while 100x−70 has a degree of 1 (x=x^{1}). 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:
Now notice that ∞ plus or minus any constant is unaffected since any number is tiny compared to ∞, so we can throw out the constant.
lim
x → −∞x^{2} 100x−70⇒ (−∞)^{2} 100·(−∞)−70⇒ ∞^{2} −100·∞− 70
Furthermore, notice that we can split ∞^{2} into ∞·∞, then cancel out one of them:∞^{2} −100·∞− 70⇒ ∞^{2} −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.)∞^{2} −100 ·∞⇒ ∞·∞ −100 ·∞⇒ ∞ −100
At this point we're done, but we need to interpret what −∞ means. Remember that ∞ is a standin 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.∞ −100⇒ − ∞ 100⇒ − ∞
 The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinitythat 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 4x^{3}−7 and the denominator of x^{2}−2x^{3} are polynomials, so we can do this here. Notice that 4x^{3}−7 has a degree of 3 (x^{3}), as does x^{2}−2x^{3}.
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:
Now we make a ratio of dividing the numerator's coefficient by the denominator's coefficient to find the limit:4x^{3}−7 ⇒ Coefficient: 4 x^{2}−2x^{3} ⇒ Coefficient: −2 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:
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.
lim
x → ∞4x^{3}−7 x^{2}−2x^{3}⇒ 4(∞)^{3}−7 (∞)^{2}−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}−7 ∞^{2}−2·∞^{3}⇒ 4·∞^{3} −2·∞^{3}4·∞^{3} −2·∞^{3}⇒ 4 −2⇒ −2
 The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinitythat 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
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 → −∞1 x= 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 ⎛
⎝1 x⎞
⎠=
lim
x → −∞cos( 0)
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
 The limit at infinity is a way of talking about what value the function approaches as x heads off towards infinitythat 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+ 10x^{2}+3 (2x−3)^{2}⎤
⎦=
lim
x → ∞⎡
⎣3x 2x−1+ 10x^{2}+3 4x^{2}−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:
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⇒ Num. degree: 1 Denom. degree: 1 ⇒ N. coefficient: 3 D. coefficient: 2
Next, the other half of the limit:
lim
x → ∞3x 2x−1= 3 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 → ∞10x^{2}+3 4x^{2}−12x+9⇒ Num. degree: 2 Denom. degree: 2 ⇒ N. coefficient: 10 D. coefficient: 4
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 → ∞10x^{2}+3 4x^{2}−12x+9= 10 4= 5 2
lim
x → ∞⎡
⎣3x 2x−1+ 10x^{2}+3 4x^{2}−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:
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:
lim
x → ∞⎡
⎣3x 2x−1+ 10x^{2}+3 4x^{2}−12x+9⎤
⎦⇒ 3·∞ 2·∞−1+ 10 ·∞^{2} + 3 4 ·∞^{2} −12 ·∞+ 9
Now we can cancel out common factors of ∞ in each fraction since they just represent the same BIG NUMBER, then simplify:3·∞ 2·∞−1+ 10 ·∞^{2} + 3 4 ·∞^{2} −12 ·∞+ 9⇒ 3 ·∞ 2 ·∞+ 10 ·∞^{2} 4 ·∞^{2}3 ·∞ 2 ·∞+ 10 ·∞^{2} 4 ·∞^{2}⇒ 3 2+ 10 4= 3 2+ 5 2= 8 2= 4
 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 infinitythat 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):
a_{n} = (n+2)! n! ·2^{n}= (n+2)(n+1)n! n! ·2^{n}= (n+2)(n+1) 2^{n}= n^{2} +3n +2 2^{n}  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 2^{n}, 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:
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:
lim
n → ∞n^{2} +3n +2 2^{n}⇒ ∞^{2} + 3·∞+ 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, n^{2} or 2^{n}? Here we might think back to the lesson on Exponential Functions and remember that 2^{n} grows extremely quickly (remember the story with the grains of rice on the chessboard from that lesson). While n^{2} gets big, it is completely outclassed by 2^{n}, 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 100^{2} is pretty large, it is absolutely tiny in comparison to 2^{100} (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} + 3·∞+ 2 2^{∞}⇒ ∞^{2} 2^{∞}∞^{2} 2^{∞}⇒ 0
 In general, the average is determined as
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 = Total amount of "stuff" Number of "objects"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:
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:Cost of producing 10 000 books = 12 000 + (10 000)(1.70) = 29 000
Thus when we print 10 000 books, the average cost is $2.90 per book.Average cost per book, printing 10 000 = 29 000 10 000= 2.90
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
Thus when we print 100 000 books, the average cost is $1.82 per book.Average cost per book, printing 100 000 = 182 000 100 000= 1.82  (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
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 haveAverage cost per book = Total cost of making books Number of books made
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:Cost of making n books = 12 000 + 1.70·n 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→ ∞:
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
Thus, as n→ ∞, the average cost per book goes to $1.70.
lim
n → ∞12 000 +1.70 ·n n= 1.70 1= 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 startup 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 startup 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.
*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 screencaptured 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
 Introduction
 Definition: Limit of a Function at Infinity
 Evaluating Limits at Infinity
 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
 What Value Does the Sequence Tend to Do in the LongRun?
 The Limit of a Sequence is Very Similar to the Limit of a Function at Infinity
 Numerical Evaluation
 Example 1
 Example 2
 Example 3
 Example 4
 Example 5
 Example 6
 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 LongRun?
 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
Precalculus with Limits Online Course
Transcription: Limits at Infinity & Limits of Sequences
Hiwelcome 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 yvalue, 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 everlarger 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 signthat means negative infinity.0127
And infinity, just on its own with nothing therewe 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 downdoes it go to some specific value l?0160
It is just that, in this case, we are talking about longterm 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 functionsin fact, the vast majority of the functions0181
that we are used to working withdo 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 valuethey grow without bound.0227
They might grow off to positive infinity; they might grow off to negative infinity (that is, verticallywhat 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)/(x^{3} + 4), h(x) = 1/x^{4},0271
j(x) = (x^{5} + 47x^{2})/(x^{3}  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 checkto 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 longterm 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 polynomial0352
some constant times x^{n}, some other constant times x^{n  1}, and working our way0358
down to a constant times x, plus some constant, and the same thing on the bottom, as well0364
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 x^{n} 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, a_{n} and b_{m}.0432
The horizontal asymptote that we get out of that is a ratio of the leading coefficients, a_{n} divided by b_{m},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 infinity0489
a horizontal asymptote is what value it approaches over the long termthey 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 a_{n} and b_{m} 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 a_{n} divided by b_{m}.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 dothere 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 downdoes 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 numbersomething largewhat 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 fraction0681
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 x0694
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, a_{1}, a_{2}, a_{3}, a_{4}, a_{5}...0747
so it is some infinite sequence that just keeps going on forever and ever and ever0752
then what we can consider is the limit as n goes to infinity of a_{n}.0756
What does our sequence go towardswhat 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 downdoes 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 farout term, it will be very large.0810
But if we look at an even fartherout 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 numericallythat 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 longterm 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 downit is going to tend to a single value in the long run?0982
As you go to those larger xvalues, 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 stepbystep 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 1it 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 x^{0}, which would only give us 1; and our bottom has x^{1}, 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 1it 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 (x^{3} + 3)/(x^{2} + 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 x^{3}.1102
But on the bottom, we have a leading exponent of squared; so we have a growth rate somewhere in the neighborhood of x^{2}.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 10^{2} and 10^{3}; 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 (8x^{4} + 3x^{2})/(2x^{4}  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
3x^{2} and 17...as we get to very large numbers, as we go farther and farther out towards infinity,1215
3x^{2} 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 8x^{4}/2x^{4}.1227
In that case, we see that the x^{4} and the x^{4} 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 herethese 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/n^{2}.1262
Once again, we see that this is n^{2}; 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 comparingwhat are the two leading coefficients?1307
5/1 = 5; and there is the limit as n goes out to infinity, the limit of the sequencewhat 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) worksit 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, thoughthe 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 farit isn't growing without bound1465
it is just bouncing between these two numberseven 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 happeningthat is why you end up having this longterm 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 downkeeps 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 2^{x}, and the limit as x goes to positive infinity of 2^{x}...1577
well, if we graph, what does 2^{x} 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 xaxis, 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, 1/2^{2}, so 1/4,1609
1/2^{3}, 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 valuesthat is, values closer and closer to 01628
as 2 to the negative number becomes very largeit 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 2^{x},1646
if we look at just the first couple of numbers, 2^{1}, and then 2^{2},1650
and then 2^{3} (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)  x^{2}/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, x^{2}/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 x^{2}1768
divided by something that is also going to contain an x^{2}.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
x^{2} doesn't change on top...divided by 4, times (x + 1)^{2}...that is just equal to x^{2} + 2x + 1.1789
So, if we multiply 4 times that expansion of (x + 1)^{2}, we get 4x^{2} + 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 x^{2}; 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
Factorialsit 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 latergoodbye!1967
1 answer
Last reply by: Professor SelhorstJones
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 âˆž? That's how I learnt it but would you find that acceptable?
1 answer
Last reply by: Professor SelhorstJones
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 SelhorstJones
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 Godsend.