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Lecture Comments (2)

1 answer

Last reply by: Professor Selhorst-Jones
Thu May 9, 2013 4:50 PM

Post by Valentina Gomez on May 7, 2013

the example problems had me cracking up! Thank you for making limits seem easy!

Finding Limits

  • The easiest limits to find are the limits of "normal" functions. Now "normal" is not a technical term, it's just supposed to mean the kind of functions we're used to dealing with. Functions where
    • It does not "break" at the point we're interested in (the function is defined and makes sense);
    • It is not piecewise and/or the point we're interested in is not at the very edge of the domain.
    Assuming the two conditions above are true for a function and the point we're interested in is the value that x approaches in the limit, it will almost always be the case that the limit for the function is the same as the value for the function.
  • The above is true because, in general, most of the functions we're used to working with don't do anything "weird". That is, they're defined everywhere, they don't have holes, and they don't jump around. Therefore the functions we're used to working with go where we expect them to go.
    If f(x) is "normal" around x= c,    then
    lim
     
    f(x)  = f(c).
    This is true even if f(x) has "weird stuff" happening somewhere else. All we care about is x→ c, so as long as the neighborhood around x=c is normal, this works.
  • Often we can't use the above because something "weird" does happen at x=c. A common weird thing is dividing by 0. In this case, we can sometimes find the value of the limit (if it exists) by canceling factors before taking the limit.

    lim
     
      KA

    KB


     
       =   
    lim
     
      A

    B


     
  • For a radical expression (one with a root), the conjugate is the same expression, but flipping the sign on one side:


     

    x2 − 3x
     

     
    − 47x       

     
          

     

    x2 − 3x
     

     
    + 47x
  • Conjugates allow us to rationalize fractions, which can sometimes help us find the value of a limit. By multiplying the top and bottom of a fraction by the appropriate conjugate, we can often find limits that would not otherwise be possible.
  • We'll discuss evaluating the limits of piecewise functions in the next lesson, Continuity and One-Sided Limits. For now though, remember: as long as you're not trying to evaluate a limit on a piecewise "breakover" (where it switches from one function piece to another), the function is probably behaving "normally" on the pieces that contain the point you care about. Thus, if it's not a breakover point, you can approach it like you're evaluating the limit of "normal" function: just plug in the appropriate value and see what comes out.

Finding Limits

Evaluate limx→ −3  2x2 + 4x −5.
  • Notice that the function in the limit (2x2 + 4x −5) is a "normal" function. There's nothing strange or weird about it: it's just a standard polynomial. It doesn't "break" anywhere, it doesn't jump around, and it just generally works as we would expect it to work.
  • When dealing with a "normal" function (or, if the whole function is not "normal", then the neighborhood around the location we're interested in is), we can almost always just plug in the location that limit is approaching. That is, if the function is "normal" at the location being approached, then

    lim
    x→ c 
    f(x)   =  f(c).
  • As we noticed in the first step, f(x) = 2x2 + 4x −5 is a "normal" function. That means we can find the limit by simply plugging in the location it is headed towards: x→ −3, so plug in c=−3:

    lim
    x→ −3 
     2x2 + 4x −5     =  2(−3)2 + 4(−3) −5

        =  2·9 −12 −5

        =  18 −17

        =  1
1
Let f(x) = {
3x+2,
x ≤ 2
x2−8,
2 < x < 4
−2x+7,
4 ≤ x
.    Evaluate limx→ 5  f(x).
  • As we noted in the previous problem and in the lesson, when we're dealing with a "normal" function, we can almost always just plug in the location that the limit is approaching. This can still be true even if the function we're working with is not "normal": all we need is that the neighborhood around the location behaves in a "normal" fashion. Thus, if the function is "normal" at the location being approached, then

    lim
    x→ c 
    f(x)   =  f(c).
  • Notice that f(x) is decidedly not "normal": it is a piecewise function, so it does something "weird"-it jumps whenever it switches pieces. The function has "breakover" points at x=2 and x=4: these make the boundary edges between the different pieces of the function. However, we also must notice that we aren't interested in those "breakover" locations! We're interested in x→ 5, so there is a small neighborhood of "normalcy" around c=5. Thus, even though f(x) is not "normal" in general, we can still plug in because the specific location is in a "normal" section.
  • Because f(x) is "normal" in the neighborhood of c=5 (the location of the limit is not a "breakover"), we can plug in. [When evaluating the function, make sure to use the appropriate piece from the piecewise function.]

    lim
    x→ 5 
     f(x)     =  f(5)

        =  −2(5)+7

        =  −10+7

        =  3
3
Evaluate limx → 4 [(3x−12)/(x−4)].
  • Begin by noticing that we can not simply plug c=4 in to the function. If we did so, we would get a 0 for the denominator, and the function would break down. Thus, the function is not "normal" at the location we're interested in.
  • Once we realize we can't simply plug in the value for the location, we should begin looking for common factors that we can cancel. Factor the top and bottom of the fraction as completely as possible. This problem is fairly simple, and it's not too hard to see that a factor of (x−4) is in the top as well:

    lim
    x → 4 
    3x−12

    x−4
        =    
    lim
    x → 4 
    3(x−4)

    x−4
  • While we can't just plug in the location of the limit, we can now cancel out the common factor:

    lim
    x → 4 
    3x−12

    x−4
        =    
    lim
    x → 4 
    3(x−4)

    x−4
        =    
    lim
    x→ 4 
    3
    After canceling the factor, we see that there is now no issue with plugging in the location of the limit. (After all, at this point there isn't even an x variable remaining, so the limit has no effect: the limit of a constant is just the constant.)

    lim
    x→ 4 
    3     =     3
    Thus, by canceling out the common factor, we were able to make the function "normal" at the limit location and find the limit by plugging in.
3
Evaluate limx → −6 [(12x+72)/(x2−36)].
  • Begin by noticing that we can not simply plug c=−6 in to the function. If we did so, we would get a 0 for the denominator, and the function would break down. Thus, the function is not "normal" at the location we're interested in.
  • Once we realize we can't simply plug in the value for the location, we should begin looking for common factors that we can cancel. Factor the top and bottom of the fraction as completely as possible.

    lim
    x → −6 
    12x+72

    x2−36
        =    
    lim
    x → −6 
    12(x+6)

    (x+6)(x−6)
  • While we can't just plug in the location of the limit, we can now cancel out the common factor:

    lim
    x → −6 
    12x+72

    x2−36
        =    
    lim
    x → −6 
    12(x+6)

    (x+6)(x−6)
        =    
    lim
    x → −6 
    12

    (x−6)
    After canceling the factor, we see that there is now no issue with plugging in the location of the limit. The limit is x → −6, so plug in c=−6:

    lim
    x → −6 
    12

    (x−6)
        =     12

    (−6−6)
        =     12

    −12
        =     −1
    Thus, by canceling out the common factor, we were able to make the function "normal" at the limit location and find the limit by plugging in.
−1
Evaluate lima → 3  [(a4−81)/(a2+2a−15)].
  • First off, don't get freaked out by the fact that the limit is based around a instead of x. The specific variable being used is unimportant: everything still works the same way. Next, notice that we can not simply plug c=3 in to the function. If we did so, we would get a 0 for the denominator, and the function would break down. Thus, the function is not "normal" at the location we're interested in.
  • Once we realize we can't simply plug in the value for the location, we should begin looking for common factors that we can cancel. Factor the top and bottom of the fraction as completely as possible. Factoring the top might be a little challenging, but notice that we can re-write a4 as (a2)2. With this in mind, we get

    lim
    a → 3 
      a4−81

    a2+2a−15
        =    
    lim
    a → 3 
      (a2−9)(a2+9)

    (a−3)(a+5)
        =    
    lim
    a → 3 
      (a−3)(a+3)(a2+9)

    (a−3)(a+5)
  • While we can't just plug in the location of the limit, we can now cancel out the common factor:

    lim
    a → 3 
      a4−81

    a2+2a−15
        =    
    lim
    a → 3 
      (a−3)(a+3)(a2+9)

    (a−3)(a+5)
        =    
    lim
    a → 3 
      (a+3)(a2+9)

    (a+5)
    After canceling the factor, we see that there is now no issue with plugging in the location of the limit. The limit is a → 3, so plug in c=3:

    lim
    a → 3 
      (a+3)(a2+9)

    (a+5)
        =     (3+3)(32+9)

    (3+5)
    Simplify from there:
    (3+3)(32+9)

    (3+5)
        =     (6)(9+9)

    8
        =     (6)(18)

    8
        =     3 ·9

    2
        =     27

    2
[27/2]
Evaluate limx → 0  [(√{2+x}−√2)/x].
  • Begin by noticing that we can not simply plug c=0 in to the function. If we did so, we would get a 0 for the denominator, and the function would break down. Thus, the function is not "normal" at the location we're interested in.
  • Once we realize we can't simply plug in the value for the location, we should begin looking for common factors that we can cancel. However, we see that the function (in its current form) cannot be factored. Notice that the function involves square roots, though: this brings up the idea of rationalization. Hopefully, if we rationalize the numerator of the fraction, we will notice a common factor that can be canceled. Let's try that. [Remember, you rationalize by multiplying top and bottom by the conjugate-the same expression, but with a flipped sign.]

    lim
    x → 0 
     


     

    2+x
     
    −√2

    x
    ·


     

    2+x
     
    +√2



     

    2+x
     
    +√2
        =    
    lim
    x → 0 
      (2+x)−2

    x(

     

    2+x
     
    +√2)
    While we could distribute the x in the denominator, it won't actually help us: our goal is to look for common factors to cancel, so unless it actually simplifies things, we want to keep things factored. Simplify the top, and we finally see the common factor we can cancel:

    lim
    x → 0 
      (2+x)−2

    x(

     

    2+x
     
    +√2)
        =    
    lim
    x → 0 
      x

    x(

     

    2+x
     
    +√2)
  • While we still can't just plug in the location of the limit, we can now cancel out the common factor:

    lim
    x → 0 
      x

    x(

     

    2+x
     
    +√2)
        =    
    lim
    x → 0 
      1

    (

     

    2+x
     
    +√2)
    After canceling the factor, we see that there is now no issue with plugging in the location of the limit. The limit is x → 0, so plug in c=0:

    lim
    x → 0 
      1



     

    2+x
     
    +√2
        =     1



     

    2+0
     
    +√2
        =     1

    √2+√2
        =     1

    2√2
    Thus, by rationalizing to find a common factor and then canceling it, we were able to make the function "normal" at the limit location and find the limit by plugging in.
[1/(2√2)]
Evaluate limk → −4  [(√{k+13}−3)/(k+4)].
  • First off, don't get freaked out by the fact that the limit is based around k instead of x. The specific variable being used is unimportant: everything still works the same way. Next, notice that we can not simply plug c=−4 in to the function. If we did so, we would get a 0 for the denominator, and the function would break down. Thus, the function is not "normal" at the location we're interested in.
  • Once we realize we can't simply plug in the value for the location, we should begin looking for common factors that we can cancel. However, we see that the function (in its current form) cannot be factored. Notice that the function involves square roots, though: this brings up the idea of rationalization. Hopefully, if we rationalize the numerator of the fraction, we will notice a common factor that can be canceled. Let's try that. [Remember, you rationalize by multiplying top and bottom by the conjugate-the same expression, but with a flipped sign.]

    lim
    k → −4 
     


     

    k+13
     
    −3

    k+4
    ·


     

    k+13
     
    +3



     

    k+13
     
    +3
        =    
    lim
    k → −4 
      (k+13) − 9

    (k+4)(

     

    k+13
     
    +3)
    While we could expand (FOIL) the two factors in the denominator, it won't actually help us: our goal is to look for common factors to cancel, so unless it actually simplifies things, we want to keep things factored. Simplify the top, and we finally see the common factor we can cancel:

    lim
    k → −4 
      (k+13) − 9

    (k+4)(

     

    k+13
     
    +3)
        =    
    lim
    k → −4 
      k+4

    (k+4)(

     

    k+13
     
    +3)
  • While we still can't just plug in the location of the limit, we can now cancel out the common factor:

    lim
    k → −4 
      k+4

    (k+4)(

     

    k+13
     
    +3)
        =    
    lim
    k → −4 
      1

    (

     

    k+13
     
    +3)
    After canceling the factor, we see that there is now no issue with plugging in the location of the limit. The limit is k → −4, so plug in c=−4:

    lim
    k → −4 
      1



     

    k+13
     
    +3
        =     1



     

    −4+13
     
    +3
        =     1

    √9 + 3
        =     1

    3+3
        =     1

    6
    Thus, by rationalizing to find a common factor and then canceling it, we were able to make the function "normal" at the limit location and find the limit by plugging in.
[1/6]
Evaluate limx→ 0  [cos(x)/(ex)].
  • Begin by noticing that the function, while unusual, is not actually "weird". While it's not something we're used to working with, it never breaks down. Why? The numerator and denominator are both defined for any x value, and the denominator can never equal 0 (because although ex can get close to 0 as x → −∞, it can never actually reach 0). Thus, we see that the function is actually "normal".
  • When dealing with a "normal" function (or, if the whole function is not "normal", then the neighborhood around the location we're interested in is), we can almost always just plug in the location that limit is approaching. That is, if the function is "normal" at the location being approached, then

    lim
    x→ c 
    f(x)   =  f(c).
  • Thus, since the function we're dealing with is actually a "normal" function (even if it's one we're not used to working with), we can just plug in c=0:

    lim
    x→ 0 
      cos(x)

    ex
        =     cos(0)

    e0
    From here, just simplify. [Remember that cos(0)=1 (from the unit circle) and e0=1 (because anything raised to the 0 equals 1).]
    cos(0)

    e0
        =     1

    1
        =     1
1
Evaluate limv→ 0  [([1/(v−2)] + [1/2])/v].
  • First off, don't get freaked out by the fact that the limit is based around v instead of x. The specific variable being used is unimportant: everything still works the same way. Next, notice that we can not simply plug c=0 in to the function. If we did so, we would get a 0 for the denominator, and the function would break down. Thus, the function is not "normal" at the location we're interested in.
  • Once we realize we can't simply plug in the value for the location, we should begin looking for common factors that we can cancel. However, we see that the function (in its current form) cannot be factored. Notice that the function currently has fractions inside of fractions, though: this brings up the idea of simplifying it so no fractions appear in the numerator or denominator. Hopefully, if we find a way to express the function without all these extra fractions, we will notice a common factor that can be canceled. Let's try that. [Remember, we can knock out these fractions by multiplying the top and bottom by each denominator.]

    lim
    v→ 0 
     
    1

    v−2
    + 1

    2

    v
    · (v−2)(2)

    (v−2)(2)
        =    
    lim
    v→ 0 
      2 + (v−2)

    v(v−2)(2)
    While we could distribute and expand all the factors in the denominator, it won't actually help us: our goal is to look for common factors to cancel, so unless it actually simplifies things, we want to keep things factored. Simplify the top, and we finally see the common factor we can cancel:

    lim
    v→ 0 
      2 + (v−2)

    v(v−2)(2)
        =    
    lim
    v→ 0 
      v

    v(v−2)(2)
  • While we still can't just plug in the location of the limit, we can now cancel out the common factor:

    lim
    v→ 0 
      v

    v(v−2)(2)
        =    
    lim
    v→ 0 
      1

    (v−2)(2)
    After canceling the factor, we see that there is now no issue with plugging in the location of the limit. The limit is v → 0, so plug in c=0:

    lim
    v→ 0 
      1

    (v−2)(2)
        =     1

    (0−2)(2)
        =     1

    (−2)(2)
        =     − 1

    4
    Thus, by "cleaning out" the nested fractions to find a common factor and then canceling it, we were able to make the function "normal" at the limit location and find the limit by plugging in.
−[1/4]
Let f(x) = √x.     Find limh→ 0  [(f(x+h) − f(x))/h].
  • First, notice that while we can't really work with [(f(x+h) − f(x))/h], we can evaluate the function with those inputs to give us something we can work with:

    lim
    h→ 0 
      f(x+h) − f(x)

    h
        =    
    lim
    h→ 0 
     


     

    x+h
     
    − √x

    h
    Second, notice that we have h→ 0, not x → 0. This means we are looking for a way to swap out h for 0. However, if we try to plug in 0 for h right now, it will break down because we'll have a denominator of 0.
  • Once we realize we can't simply plug in the value for the location, we should begin looking for common factors that we can cancel. However, we see that the function (in its current form) cannot be factored. Notice that the function involves square roots, though: this brings up the idea of rationalization. Hopefully, if we rationalize the numerator of the fraction, we will notice a common factor that can be canceled. Let's try that. [Remember, you rationalize by multiplying top and bottom by the conjugate-the same expression, but with a flipped sign.]

    lim
    h→ 0 
     


     

    x+h
     
    − √x

    h
    ·


     

    x+h
     
    +√x



     

    x+h
     
    +√x
        =    
    lim
    h→ 0 
      (x+h) − x

    h(

     

    x+h
     
    +√x)
    While we could distribute and expand the factors in the denominator, it won't actually help us: our goal is to look for common factors to cancel, so unless it actually simplifies things, we want to keep things factored. Simplify the top, and we finally see the common factor we can cancel:

    lim
    h→ 0 
      (x+h) − x

    h(

     

    x+h
     
    +√x)
        =    
    lim
    h→ 0 
      h

    h(

     

    x+h
     
    +√x)
  • While we still can't just plug in the location of the limit, we can now cancel out the common factor:

    lim
    h→ 0 
      h

    h(

     

    x+h
     
    +√x)
        =    
    lim
    h→ 0 
      1

    (

     

    x+h
     
    +√x)
    After canceling the factor, we see that there is now no issue with plugging in the location of the limit. The limit is h → 0, so plug in c=0:

    lim
    h→ 0 
      1



     

    x+h
     
    +√x
        =     1



     

    x+0
     
    +√x
    Finally, simplify the expression as much as you can. It won't come out as a number, but that's okay: we weren't told a value to use for x, so x should appear in the final result.
    1

    √x+√x
        =     1

    2√x
[1/(2√x)] [While the problem makes no mention of it, the limit expression limh→ 0  [(f(x+h) − f(x))/h] is a very special thing in calculus. This problem foreshadows the idea of the derivative, which, while currently meaningless, will turn out to be an incredibly important idea for later mathematics courses (and science, economic, and engineering courses!). You'll get a glimpse of the concept in the second-to-last lesson of the course, Instantaneous Slope & Tangents (Derivatives).]

*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

Finding Limits

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 0:08
  • Method - 'Normal' Functions 2:04
    • The Easiest Limits to Find
    • It Does Not 'Break'
    • It Is Not Piecewise
  • Method - 'Normal' Functions, Example 3:38
  • Method - 'Normal' Functions, cont. 4:54
    • The Functions We're Used to Working With Go Where We Expect Them To Go
    • A Limit is About Figuring Out Where a Function is 'Headed'
  • Method - Canceling Factors 7:18
    • One Weird Thing That Often Happens is Dividing By 0
    • Method - Canceling Factors, cont.
    • Notice That The Two Functions Are Identical With the Exception of x=0
    • Method - Canceling Factors, cont.
    • Example
  • Method - Rationalization 12:04
    • Rationalizing a Portion of Some Fraction
    • Conjugate
    • Method - Rationalization, cont.
    • Example
  • Method - Piecewise 16:28
    • The Limits of Piecewise Functions
  • Example 1 17:42
  • Example 2 18:44
  • Example 3 20:20
  • Example 4 22:24
  • Example 5 24:24
  • Example 6 27:12

Transcription: Finding Limits

Hi--welcome back to Educator.com.0000

Today, we are going to talk about finding limits.0002

We often need to find the precise value that a limit will produce.0004

However, the methods we saw when we first introduced limits (that is, graphing it or a table of values for the function) are not precise.0007

They can give us a good idea of what the limit will be, but they don't give us certainty.0015

They don't let us know that it will be exactly something.0020

Likewise, the formal (ε,δ) definition of a limit that we talked about in the last lesson--0022

and it is totally fine if you do not know it; that was a completely optional lesson, only if you are really interested in math,0027

and wanted to find out more about stuff that is going to come later in a few years if you keep studying math;0033

it is totally fine if you didn't do it--but if you did, that is still not really going to help us find limits.0038

It allows us to formally prove that this limit has to be here.0043

And it is the deeper mechanics of what is going on "under the hood" for how a limit works.0047

But it doesn't let us find limits; it doesn't make finding them easier; it is just about proving limits.0051

In this lesson, we will see various methods to find the precise value; that is what this lesson will be about.0056

The basic idea that we are going to see with all of these methods is to transform the function into something0061

that works pretty much the exact same way, that we will just be able to plug in a value for the x0066

in this equivalent version, and we will be able to churn out some value for what the limit will come out to be.0072

Now, before you watch this, make sure that you are already familiar with the concept of a limit.0077

You really want to have a good understanding of how a limit works.0083

We will be working out how to get numbers in this lesson.0086

But if you don't actually understand what this stuff means, it is all going to fall apart really quickly.0088

So, it is really important that you understand how a limit works before you watch this.0092

If you don't already have a good understanding of how limits work, check out the lesson two lessons back,0096

Idea of a Limit, where we will explain and get an idea of what a limit is about.0101

And that way, we will have some meaning for how we actually get precise values.0105

You can figure out the precise values without really understanding what is going on.0110

But that will fall apart really, really quickly; and you might as well just have a nice foundation to work from there.0114

It won't take that long to get an idea of what is going on.0119

All right, first, the easiest limits to find are limits for "normal functions."0121

And now, that is in quotes, because normal is not a technical term.0127

What I just mean here is...it is supposed to mean the kind of functions that we are used to dealing with,0131

the sort of thing that we use most often, functions where it does not break at the point we are interested in--0135

that is to say, the function is defined and makes sense; it doesn't suddenly break down when we get to the place that we really care about.0142

And it is not piecewise, and/or the point that we are interested in is not at the very edge of the domain.0148

The point that we care about, whatever x we are going to (we are going to some x going to c)...0154

whatever c we are going towards, it is not going to be at the very edge of the domain, or where we split on some piecewise function.0159

So, as long as that location makes sense--everything in that area makes sense--0166

we know how to use the function in that area around that place, and it isn't piecewise--0169

there aren't different parts of it, and it is not the very edge of the domain,0174

the very starting or very last value for it--as long as those aren't the case, we will be able to do it really easily.0177

If these two conditions are met, where it isn't breaking down and it is not piecewise,0184

and the point isn't at the edge of the domain, then it is really easy to figure out what the value is going to be.0189

So, if the point we are interested in is the value that x approaches in the limit--0195

that is to say, x goes to c, and c is the place where things aren't breaking down,0200

and c is not at a piecewise break-over, and it is not at the very edge of the domain,0204

then it is almost always going to be the case that the limit as x goes to c of f(x)0208

is as simple as just plugging c in for x and getting f(c).0212

So, let's get an example first, to see how we could use this.0218

If we looked at the limit as x goes to 2 of 1/x2, what would that end up being?0220

Well, first notice: while 1/x2 breaks down (it isn't defined, that is to say) at x = 0, that is here.0225

But we don't care about x = 0; that is not the area that we are interested in.0232

We care about x going to 2; so if x is going to 2, if we look in that area over here, that region, that is totally fine; it makes perfect sense.0236

1/x2 works fine in the region around x going to 2.0245

As x goes to 2, well, all of this stuff makes sense; we can see it.0250

It clearly just maps to values, and it is perfectly reasonable.0254

Second, 1/x2 is not a piecewise function; we don't have to worry about that.0258

And x = 2 is not at the edge of the domain (and the domain for 1/x2 is every x, with the exception of x equaling 0; it is all x not equal to 0).0261

So, we are not at the edge of a domain; it is not piecewise; and it makes sense in the area we are looking for.0270

It makes total sense in here; so since it does all of those, it means that we can just plug in the value that we are going to.0275

Because of these two conditions, limit as x goes to 2, we just plug it in for x,0282

and we have 1/22, which simplifies to 1/4, and there is our limit.0286

Why can we do this--what is the reason that we can get away with doing this?0291

Well, in general, most of the functions we are used to working with don't do anything weird.0294

They aren't strange in any way; and what I mean by not being weird is that they are defined everywhere; they don't have holes; they don't jump around.0299

They are defined anywhere that we might be interested in looking; they don't have any holes in them; and they don't jump around.0311

They work in a pretty reasonable way; they work normally--they are not weird.0317

What this all means is that the functions we are used to working with, the functions that we normally work with, go where we expect them to go.0322

Since a limit is about figuring out where a function is headed (a limit is about what our expectation is0330

for this function), and a normal function has our expectations fulfilled (what we expect from the function0336

is what we get out of the function), that means that we can evaluate normal functions at the location the limit approaches.0343

Our expectation, the limit as x goes to c of f(x), what we expect to end up landing on in our journey,0349

what we expect as we come in, ends up being what it actually is--what it is at the location.0356

So, if it is normal, if f(x) isn't doing something weird, then what we expect ends up being what we actually get.0363

So, the limit as x goes to c...well, we can just plug that in for f(x),0371

and we have that f(c) will end up being the limit, any time that we are dealing with a normal function.0374

In fact, this is true even if f(x) does have weird stuff, but as long as it happens somewhere else.0381

All we care about is x going to c; so as long as the neighborhood around x going to c is normal,0387

and the weird stuff happens off somewhere else--it isn't happening directly on top of that c--0392

then the weird stuff...we don't care about it, because the region we care about being normal is the region around x = c.0397

As long as the neighborhood around x = c is normal, as long as around x = c does this,0403

then we will end up being able to just plug into that, just fine.0409

So, if there is weird stuff, that can be OK, as long as it is far enough away.0413

When we looked at 1/x2, there was weird stuff at x = 0, but we didn't care about x = 0.0416

We cared about x going to 2; so as long as the weird stuff isn't right on top of where we are going to,0421

we can just plug in our c, and that will tell us what the limit will come out to be,0426

if it is this fairly normal function that we have been working with for years and years.0431

Of course, sometimes x goes to c, but something weird does happen at x going to c.0436

When we are at x = c, it is a weird place.0441

One weird thing that often happens is dividing by 0; you are not defined when you divide by 0.0445

So, consider the function and limit below: f(x) = 2x/x.0450

Well, we can see it graphed here; it makes perfect sense, up until we try to plug in x = 0, at which point the function breaks down.0455

But the limit is pretty clear; it is going to 2 the entire time, so what is it headed towards?0462

It is headed towards 2; that is what we get out of the limit.0466

So, of course, we can clearly see that the limit in this case exists, and it is 2.0469

However, let's also notice that with the exception of x = 0, everywhere other than actually at x = 0,0474

where the weird thing happens, the function f(x) = 2x/x is just equivalent to if we had canceled out those x's and gotten g(x) = 2.0481

Notice: f(x) = 2x/x and g(x) = 2...these two functions are identical, with the exception at x = 0.0492

Everywhere other than x = 0, these two are totally the same.0503

f(x) = 2x/x and g(x) = 2 behave exactly the same, with an exception at x = 0.0509

However, since we are looking at the limit as x goes to 0, we don't care about x = 0.0516

It is about the journey, not the destination; so if it is x to 0, the destination we don't care about is at x = 0.0523

So, the weird thing here at x = 0...we might as well forget about it.0529

That means, since we don't care about what happens at the weird place, and g(x) = 2 is exactly the same0533

everywhere but the weird place, that means that we can use g(x) to evaluate the limit.0540

And we can just plug in g(0) to get 2.0545

So, g in general works just the same as f; g is just the same as f.0548

It works the same everywhere, with the exception of this one weird little point.0555

However, since we are looking at a limit going to that one weird little point, we don't actually care about the weird little point.0559

A single point doesn't matter, because the limit is about the journey towards that point.0564

So, g(x) = 2 behaves the exact same for the journey portion.0568

The journey towards behaves the exact same, whether we are using f or g, which means that g(x) is what we can use for figuring out the limit.0574

And then, by the same logic we were just talking about previously, since g(x) is totally normal at 2,0584

we can just plug in there, and we can get the answer from g(x).0588

Now, how did we find g(x)?--by canceling common factors.0591

This logic works in general; if we have some function given as a fraction, we can cancel out factors between top and bottom,0598

because a single point...if we cancel a factor, the only thing that can possibly happen0607

that would be bad is that we would cause one point to change around, like with f(x), x = 0 did not exist;0611

but with g(x), x = 0 did exist; so we caused a problem for specifically one point.0617

But a single point has no effect on a limit, because with a limit, it is about the journey, not a single point that is our destination.0622

So, since a journey is made up of a multitude of points, taking out a single point,0634

changing a single point, doesn't actually have an effect on where we end up landing for our limit.0638

That means, any time we have common factors for a limit, we can cancel out common factors0643

and just get what it would be without those common factors for the limit.0647

We will have changed the function that we are using, but the limits will be equivalent.0651

So, here is an example: we have limit as x goes to 3 of (x - 3)(x + 2)/(x - 3).0656

Well, that means that, if we were to plug x = 3 into this, we would have 0/0; 3 - 3 turns to 0; 3 - 3 on the bottom turns to 0; so we would get 0/0.0661

And so, we can't do that; it goes crazy; it is weird there.0671

But we can cancel out x - 3 and cancel out x - 3, just like we canceled out the k's here; and we get some other something over something.0674

We get what remained, a and b; in this case, we have x + 2 divided by 1 now.0682

So now it is the limit as x goes to 3 of x + 2; that is effectively going to work the same.0687

x + 2 is pretty much equivalent to x - 3 times x + 2 over x - 3, with the exception of x = 3.0692

But since we don't care about that, since that is where we are headed towards, we can end up using that limit instead.0700

So, we now plug in x going to 3 into x + 2.0704

Well, x + 2 is perfectly normal at x = 3; nothing weird happens there.0707

So, since nothing weird happens there, it works normally; we can just plug our value in there.0712

We plug in 3; 3 + 2...we get 5; our answer is 5 for the limit.0716

A similar idea to canceling out factors is rationalization: rationalizing a portion of some fraction.0722

If we have a radical that is in our way, and it is part of a fraction, we can change it into a non-radical0728

by multiplying the numerator and denominator by the same thing.0735

What do we get rid of a radical with--how do we rationalize an expression?0738

We rationalize an expression that contains a radical by multiplying by its conjugate.0743

That is the same expression, but now with a negative on one side.0748

For example, if we have √x + 2, well, that is a radical and some non-radical thing.0752

If we go through the conjugate process, we get √ - 2; plus switches to negative.0756

If we have √(x2 - 3x) - 47x, some radical minus something that is not in a radical,0762

then its conjugate is √(x2 - 3x) + 47x.0768

So notice: plus, if we are going through a conjugate, becomes minus; and minus becomes plus.0773

We just swap the sign on one of the portions, and that is how we get the conjugate.0781

Now, if we multiply a radical expression by its conjugate, we will end up canceling out the radicals.0785

And we will see how that works in just a moment.0790

Since multiplying a fraction on the top and bottom by the same thing gives an equivalent expression--0793

if I have a fraction, I can multiply it by 5/5, because 5/5 is just the same thing as 1, so it is still equivalent--0798

we can figure out limits by doing this thing.0804

And we can trust that this works, that this doesn't introduce any issues, by the exact same logic that we use to cancel out factors.0806

If we were to multiply by something over something, it could introduce one slight issue;0813

but it would only introduce a slight change to the function at one point.0818

But because it is a limit, we don't care about single points on their own; we only care about continuums of points.0821

So, that single point being changed is not really an issue; the limit will still work the same.0826

So, if we have the limit as √(x + 4) - 2, over x, as x goes to 0,0831

well, we would want to multiply this by the conjugate, √(x + 4)...but it was a minus previously,0838

so now it swaps to a + 2, and it will have to be divided by the same thing,0844

because we can only multiply by 1 effectively, which means the same thing on top and bottom: + 2.0848

And we multiply by that, and we start working things out.0854

What do we get out of that? Well, limit as x goes to 0 of √(x + 4) - 2, over x, times √(x + 4) + 2,0857

over √(x + 4) + 2...well, the top here is now going to become x + 4 - 4.0864

And if we are not quite sure how we see that, let's look at √(x + 4) - 2 times √(x + 4) - 2.0869

Well, what is...oops, I should not have put that radical over the whole thing; the radical ends there.0878

What is √(x + 4) times √(x + 4)?0883

Well, the square root of thing times the square root of same thing always just lets out the thing on its own.0885

The square root of smiley-face times the square root of smiley-face becomes smiley-face.0890

So, √(x + 4) times √(x + 4) becomes x + 4.0894

So, x + 4 is what we get out of that; and now we have √(x + 4) times -2...0898

oops, that shouldn't be -2; it should be plus, because it is a conjugate; I'm sorry about that.0905

√(x + 4) times positive 2 is 2√(x + 4): and then -2 times √(x + 4) is -2√(x + 4).0909

So, we have positive 2 √(x + 4) and -2√(x + 4); those two things cancel each other out.0916

Positive 2 and negative 2 cancel each other out; so the middle part disappears.0922

And now, it is -2 times +2; that becomes -4; so that is where we get x + 4 - 4.0927

On the bottom, it is x times this thing over here; so we just put in the quantity, because we will have some convenient canceling happening very soon.0934

On our top, we have x + 4 - 4; so plus 4 and minus 4 cancel each other out; we are left with just x on top;0943

it is still x times √(x + 4) + 2 on the bottom.0948

But now, we say, "We have x on top and x on bottom," and now we have the limit as x goes to 0 of 1/(radic;(x + 4) + 2).0952

At this point, we see that if we plug in 0...does anything weird happen?0959

Well, 1/(√(0 + 4) + 2)...we aren't dividing by 0 anymore, so it effectively works as a normal function.0964

We don't have any weird thing happening, so we plug in for our x at this point.0971

The square root of 0 + 4, plus 2 is in our denominator: 1 over...√4 is 2, so we have 1 over 2 + 2, which gets us 1/4.0975

And that is how we get to our answer for that limit.0985

We will discuss evaluating the limits of piecewise functions.0988

We haven't talked about piecewise functions yet, because we will be talking about that in the next lesson, Continuity and One-Sided Limits.0991

We will want a couple of little new ideas before we talk about piecewise functions.0997

That is why we are saving them for the next lesson.1001

For now, though, remember: as long as you are not trying to evaluate--as long as it is not trying1002

to evaluate the limit at some piecewise break-over, where it switches from one piece to another piece,1007

the function is probably going to be behaving normally on the pieces that contain the point you care about.1014

It might have a piecewise here and a piecewise here, and then it suddenly switches over.1018

But as long as we are over here in the normal area, or we are over here in the normal area, nothing weird happens.1021

So, if it is not at a break-over point, then that means that, since it is behaving normally,1027

you can approach it the same way as you do when just dealing with the limit of a normal function.1032

Just plug in the appropriate value and see what comes out.1037

Plug in the value that you are going towards and see what comes out.1040

However, if you do need to evaluate a break-over point, where you have to be talking1043

about where it switches from one to the next, check out the next lesson,1048

because we will see how that idea works specifically in the next lesson.1051

All right, we are ready for some examples.1055

The first one: Evaluate the limit as x goes to 2 of x4 - 3x2 + 4x - 10.1057

Our first question that we want to ask ourselves is if x4 - 3x2 + 4x - 10 is normal.1062

Yes, there is nothing weird that happened in that; that is just a polynomial--we are used to that sort of thing.1066

So, it is normal; if it is normal, then that means that we can just plug in our value for each of the x.1071

So, we plug in our 2, because we know that the limit of what it gets is...well, we don't have to plug x into 10...1078

we know that the limit as what we are going to get out of this is just the same thing as what the function would be there.1084

What we expect is what we get when we are dealing with a normal function.1089

So, we plug in 2; we now have 24 - 3(2)2 + 4(2) - 10.1094

24 is 16; minus...3(2)2...22 is 4; 3 times 4 is 12,1103

plus...4 times 2 is 8; minus 10; 16 + 8 gets us 24; minus 12 minus 10 gets us -22; we put those together, and we now have 2--done!1109

The next one: let f(x) equal x2 - 3 when x is less than or equal to 2 and 5x + 2 when x is greater than 2.1120

First, let's just see, really quickly, what this looks like.1127

Here is a rough picture of what it looks like; I will do that with blue here; so x2 - 3...that is basically like a parabola.1131

It has just been lowered by 3 from a normal standard parabola.1139

And it goes until x ≤ 2, at which point it stops here, at 2.1143

And then, after that, we are at x > 2, so we switch over to 5x + 2 when x is greater than 2.1147

It starts here, and then it goes off like this.1153

And that is what we end up seeing.1155

The question here is: if we are going to evaluate the limit as x goes to 1 of f(x)...oh, no, it is a piecewise function!1157

Well, yes; but we are clearly contained within x ≤ 2.1165

The area we care about is this area here; that is far enough from something weird happening.1169

Something weird does end up happening over a little bit further to the right.1174

But we don't care about that, because in the neighborhood we are interested in (that is x going to 1),1178

we are definitely less than or equal to 2, if we are close enough to 1.1183

So, since being close enough to 1 means nothing weird happens, all we are dealing with is x2 - 3.1187

So, we are effectively normal, because we only have to consider the part of the piecewise function that we are inside of.1194

The part of the piecewise function that we are inside of is x2 - 3.1201

So, we can just plug our 1 into x2 - 3; we plug in our 1; 12 - 3; 1 - 3 gets us -2, and there is our limit.1204

The next example: Evaluate the limit as x goes to -3 of (x2 + 3x)/(x2 - 9).1216

Our first question is, "Is it normal?" Well, if we plug in -3, what do we get on the top?1221

Well, we will get 9 - 9; so that is 0; and on the bottom, we will get x2 - 9,1226

so that will be positive 9, minus 9; oh, we get 0/0; so that means it is not normal.1231

Oh, no! But what do we do as soon as we are not normal?1237

We start thinking, "All right, well, what else can we do?"1241

We have the possibility of canceling factors; so what we want to do now is think, "Is there a way for us to cancel out factors?"1243

Can we cancel out factors? Well, we have x2 + 3x...1251

so I will write limit as x goes to -3...technically, we really should have the limit at every step.1254

If you end up really not feeling like writing out the whole thing, at least write limit, so that we know1259

that we are still dealing with the limit; and we will plug in something later.1263

(x2 + 3x)/(x2 - 9): well, limit as x goes to -3...how can we change the top?1266

How can we factor the top? Well, we could pull out an x, and we would have x(x + 3) on top.1277

How can we factor the bottom, x2 - 9? Oh, that is just the same thing as (x + 3)(x - 3).1282

Great; we can cancel factors; we cancel x + 3 and cancel x + 3.1289

And that is fine, because we are just working with a limit.1295

Our limit is now the same thing as x goes to -3 of x on top, divided by x - 3 on the bottom.1298

Now, we ask ourselves, "If we plug in -3, does anything weird and disastrous happen?"1307

-3 on top; -3 on the bottom; we don't get 0/0; we don't even get dividing by 0 once.1311

We are totally fine; so we plug in, because now it is effectively a normal function, since something weird isn't happening.1317

So, we have -3/(-3 - 3); that gets us -3/-6; the negatives cancel; 3/6 is 1/2, and so we get 1/2 as the answer to our limit.1324

Great; the next one: Evaluate the limit as x goes to 0 of tan(x)/sin(x).1337

The first question that we ask ourselves is, "Is this normal?"1342

Well, if we plug in 0, tan(0) is 0; sin(0) is 0; oh, that means it is 0/0; so it is not normal.1347

But if it is not normal, the first thing we try is asking ourselves, "Can we cancel factors?"1357

So, if we can cancel factors, we are good; how can we cancel factors?1363

Limit as x goes to 0...how can we change tan(x)?1367

Well, tan(x)...remember, that is just the same thing as sin(x)/cos(x).1372

And any time we don't have just sine and cosine, and we are dealing with trigonometric stuff,1377

it normally helps to put it just in terms of sine and cosine; so we have sin(x)/cos(x), all divided by sin(x).1380

Oh, OK; well, now we see that we can start canceling stuff; limit as x goes to 0...1390

well, we could rewrite this as sin(x)/cos(x); we could just cancel out the sin(x) if we see that directly.1395

But if we find it difficult to divide fractions and fractions, we could think of this as divided by sin(x),1401

which means that that is the same thing as limit as x goes to 0 of sin(x)/cos(x), times 1/sin(x).1407

If you divide, it flips to a fraction, which we could have also done by just breaking out the fraction of the sin(x) on the bottom to the side.1416

So now, we see that there is sin(x) on the bottom and sin(x) on the top.1422

We cancel some stuff out; limit as x goes to 0...now we have 1/cos(x).1426

Since we managed to cancel some stuff out, let's ask ourselves, "If we plug in 0, is it weird? Is it normal? What happens?"1432

Well, x goes into 0 for cos(x); cos(0) is 1; 1/1 is totally not weird anymore.1438

So, that means we can now swap in our 0; 1/cos(0)...cos(0) is 1; we have 1/1, so our limit comes out to be 1.1445

OK, the next one: Evaluate the limit as x goes to 4 of (2 - √x)/(4 - x).1460

The first thing we ask ourselves is, "Is it normal?"1466

Well, if we plug in 4 on the top, we will get 2 - √4, so that will come out to be 0,1468

divided by 4 - 4...even worse...dividing by 0...so 0/0 is not normal--no! No!1472

It is not normal, so the next thing we ask ourselves is if we can cancel factors.1480

Well, 2 - √x...4 - x...we might be able to figure out a way to cancel factors, but not easily.1484

Canceling factors...we might be able to figure out a way; we could figure out a way;1490

but let's say that we don't want to figure out canceling factors; canceling factors is not easy.1496

So, there is a radical; what was the trick we learned for dealing with radicals?1501

We rationalize; we move on to the next trick in our selection; we rationalize.1505

What do we do? We have the limit as x goes to 4 of 2 - √x, over 4 - x.1514

How do we rationalize? We multiply by the conjugate on the top and the bottom.1523

2 - √x: its conjugate is 2 + √x; we could also put a negative on the 2, but it doesn't really matter which side gets the negative.1526

2 + √x; 2 + √x; great--multiply our tops together, and multiply our bottoms together.1534

2 - √x times 2 + √x: these are now factors with parentheses around them, because we have to have distribution going on.1542

So, 2 times 2 is 4; 2 times √x is 2√x; -√x times 2 is +2√x and -2√x; they cancel each other out;1550

-√x times +√x becomes -x (√x times √x always comes out to be x;1559

√smiley-face times √smiley-face always comes out to be smiley-face; root(root) cancels the roots,1565

as long as it is the same thing underneath it).1569

4 - x times 2 + √x: well, we could multiply them together--but 4 - x is what we have on the top right now.1571

We are basically working towards canceling out factors; so 2 + √x...1577

at this point, we are now going to be able to cancel out factors.1582

We have the 4 - x on the bottom and the 4 - x on the top; they cancel each other out.1587

And now, I have the limit as x goes to 4 of 1/(2 + √x).1592

Great; so, if we were to plug 4 into this, would something weird happen?1601

1/(2 + √4)...no: we don't have 0 showing up; we aren't dividing by 0.1605

It basically works like a normal function; it does work like a normal function.1609

Nothing weird is going on there; so that means we can just plug in our value.1613

So, 1 over 2 plus...plugging in 4 for our x...1 over 2 + 2 (the square root of 4 is 2)...and 1/4 is our answer; nice.1617

The final example: Evaluate the limit as x goes to 0 of 1/(x + 3) - 1/3, all divided by x.1632

The first question we ask ourselves is, "Is it a normal function?"1638

So, is it normal? Well...you can guess by my hint: no.1641

If we plugged in 0, it would be 1/3 - 1/3, so 0 on top, divided by 0 on bottom...no, it is definitely not normal.1646

So, it is not normal; oh, no, what are we going to do?1654

Well, the next thing we ask ourselves is, "Can we cancel factors?"1658

Can we cancel? Not easily: 1/(x + 3) - 1/3...I really don't see any easy ways to make cancellation show up there.1663

So, we are probably not going to be able to cancel, at least not easily.1672

So, our next question is...last time we asked ourselves, "Can we rationalize?"1675

Well, there are no radicals here, so we can't rationalize.1678

But we can take a hint from the idea of rationalization.1681

The idea of rationalization was to multiply the top and the bottom by something that makes some part not weird anymore,1684

not as strange to deal with, so that hopefully, we can get cancellation to appear later.1689

Well, what would make the top...the thing that is really strange about this is that we have fraction over fraction.1693

We don't like fractions and fractions; so how can we get rid of some of those fractions by multiplying?1699

Well, the easiest way to get rid of the denominator in the top is to just multiply by the denominators in the top.1706

If we multiply...I will rewrite the thing out...limit as x goes to 0 of 1/(x + 3) - 1/3, all over x:1712

well, what would get rid of the x + 3? x + 3 would get rid of the denominator of x + 3.1723

What would get rid of 1/3? Well, multiply by 3.1730

We can get rid of both of those denominators by multiplying that whole top by (x + 3) times 3.1733

That will cancel out each of the denominators as we work through it.1738

And remember: it is always going to multiply the whole thing; when we multiply, we multiply the quantity, because we have to have distribution showing up.1740

And on the bottom, we will have to have the same thing, because otherwise we are changing the expression; we can't change the expression.1747

x + 3, times 3, over x + 3, times 3; great; our limit continues...limit as x goes to 0...1754

What do we get on the top? Well, (x + 3) times 1/(x + 3)...those cancel out, and we are left with just the 3 left over.1760

So, (x + 3) times 3, on 1/(x + 3)...the (x + 3)s cancel out; we are left with just 3.1767

Minus...when (x + 3) times 3 hits 1/3, well, the (x + 3) doesn't do anything; but the times 3 cancels out, so we are left with minus (x + 3).1774

All right, now we could expand the bottom, but that won't actually help us.1785

One of the ideas that we are going to hopefully manage to get to is to figure out a way to cancel things.1789

We couldn't cancel things easily by factoring; but hopefully we will still manage to cancel something at some point later on.1793

We don't want to expand factors; we want to actually keep up this process of keeping things in factors.1799

So, let's not put anything together; we will have it as x times x + 3 times 3.1805

At this point, we see x + 3 on top and x + 3 on the bottom, but we have to be careful; don't cancel stuff.1810

We can't cancel, because there is still a subtraction sign on the top; we have to have the whole factor.1814

We keep working to simplify: limit as x goes to 0...3 - (x + 3)...well, the 3's will cancel out,1818

and we will be left with just -x on the top, divided by x times (x + 3) times 3; great.1825

Limit as x goes to 0 of -x/x(x + 3)(3); at this point, we say, "We can cancel some stuff!"1838

This x and this x cancel, and we are left with the limit as x goes to 0 of -1 now (because it just canceled out the x,1844

not also the negative), times (x + 3), times 3.1852

Now, we ask ourselves, "Now that we have managed to cancel something, if we were to plug in a number, would we have something weird happen?"1856

Would it be normal now? If we plug in 0, we get -1/(0 + 3)(3); it doesn't look like we are going to be having dividing-by-zero issues anymore.1862

It isn't weird anymore, so we can just plug in; this is equal to the -1, over (0 + 3) times 3.1869

Once it is not weird, we can plug in, because now it is effectively normal.1879

And when it is effectively a normal function, you can just plug into it with your limit.1882

-1/3(3) gets us -1/9; and there is our answer--great.1887

All right, at this point, we have a really good understanding for how to figure out how limits work.1896

The basic idea is, "All right, I have a limit that is normal and doesn't have anything weird happen."1900

That is easy--just plug in something and crank it out--see what number you get out.1905

That is what the limit is, because "normal" means that your expectations will be met.1908

If it is not normal, if there is something weird happening, you try to manipulate things.1912

You either pull out factors or multiply the top and the bottom; you do something where you are allowed to cancel factors later on.1916

And then, you check and see, "OK, now that I have canceled out the factors, is it possible for me to plug things in and have it be normal, effectively?"1922

Can I now plug in (now that there isn't, hopefully, a weird thing happening)?1929

Sometimes there will still be weird things happening; in all of the examples we saw here, we canceled out anything that would cause weird stuff to happen.1932

But sometimes, you will end up still having weird stuff, no matter what you manage to cancel out.1939

And in that case, it can help to check a graph and think, "Oh, I see: it is going to go out to infinity," or something like that.1943

And you will see that it is never going to work.1947

But a lot of the time, you can cancel stuff out, and you can say, "Oh, OK, now it is effectively behaving like a normal function;1949

so I can plug in the x-value that I am going towards and just crank out an answer."1954

All right, we will see you at Educator.com later--goodbye!1958