Exercises for section 2.1

Worked Example

Prove the following limit:

               2 (x² - 4)
     lim                   =  8                                  eq. 2.1x-1
    x → 2       (x - 2)

using the delta-epsilon method.

Clearly we cannot evaluate this function at x = 2 because that would make for a zero denominator.

We recall from algebra that x² - 4 is the difference of squares and can therefore be readily factored.

   x² - 4   =   (x + 2) (x - 2)

Let's give the function we are trying to find the limit of a name. Let's call it f(x). So we have:

                2 (x² - 4)       2 (x + 2) (x - 2)
     f(x)   =                =                                  eq. 2.1x-2
                  (x - 2)             (x - 2)

Clearly at all values of x except x = 2 we get a cancellation, and this is the same as:

   f(x) = 2 (x + 2)                                             eq. 2.1x-3

So the above holds for any value of x except 2. That means we can evaluate f(x) using this expression and get the right answer provided x is never equal to 2. So if we prove that:

     lim    2 (x + 2)   =   8                                   eq. 2.1x-4
    x → 2

then we have solved the problem. Do you understand why? Remember that all the expressions we have made for f(x) are identical functions for any value of x besides 2. Taking the limit of f(x) as x goes toward 2 means never having to evaluate f(x) at x = 2. So at all the places we do have to evaluate it, all the forms of f(x) are indeed identical, including f(x) = 2 (x + 2).

Now we get down to the delta-epsilon part of the proof. We have to arrange a scheme by which you can tell me how close f(x) has to be to 8 (i.e. you give me an ε) and based upon that can tell you how close x has to be to 2 to make it true (i.e. I can give you a δ that makes it true).

Well, let's just set up an equation that shows what happens when we use an x that is within δ of 2.

   f(2 ± δ)  =  2 ( (2 ± δ) + 2)                                 eq. 2.1x-5

Now remembering that δ is always greater than zero (and more importantly it never is zero), we can see that in the above, we still never have to evaluate f(x) at the forbidden value. So we just multiply out the above expression:

   f(2 ± δ)   =   8  ±  2 δ                                      eq. 2.1x-6

The requirement is that we have to be able to choose δ so that the value above is no farther from the limit (which in this case is 8) than the ε that you might give me, no matter how small an ε you do give me. So by giving me an ε, you are telling me to make it so that:

   |f(2 ± δ)  -  8|   ≤  ε                                        eq. 2.1x-7

But we can get an expression for what's inside the absolute value brackets from stuff we have already done. Just take equation 2.1x-6 and subtract 8 from both sides. If you substitute that in for 2 ± δ, you get:

   |± 2 δ|   ≤   ε                                                eq. 2.1x-8

and since the absolute value brackets make the ± sign moot, we have simply:

   2 δ   ≤   ε                                                    eq. 2.1x-9

           ε
   δ    ≤                                                        eq. 2.1x-10
           2

So, if you hit me with any ε, all I have to tell you is to try a δ that is less than or equal to half of your ε. In other words we have established a scheme that turns ε's into δ's, and the scheme always gives you a δ that makes the function come within ε of 8. And that means we're done with the proof.

Another Worked Example

It seems that many instructors would like you to recall as much of the material you learned in algebra and apply it to your calculus exercises. They are not trying to torture you with this, but rather they are trying to teach you to use your mathematical "vocabulary." A typical type of problem that most of them assign is one that requires you to use the difference of squares. So here is a problem that requires you to do just that: Find the limit of

             (x - 3)
    lim                                                           eq. 2.1x-11
   x → 3   √x + 1 - 2

Notice that the denominator is zero at x = 3, so you need to take the limit somehow. At first it might seem beyond your algebraic abilities. But whenever you see a square root plus or minus something, alway think difference of squares. Multiply top and bottom by the complement of the square root sum. That is, if the square root sum is √a + b, then multiply top and bottom by √a - b. Likewise if the square root sum is √a - b, then multiply top and bottom by √a + b. In this case you would multiply numerator and denominator by √x + 1 + 2

                _____
              (√x + 1 + 2)(x - 3)
    lim                                                           eq. 2.1x12a
   x → 3   (√x + 1 + 2)(√x + 1 - 2)

Do you remember from algebra that whenever you have (a + b)(a - b) that it is always the same as a² - b²? That is precisely what we have in the denominator of eq. 2.1x12a. So we go ahead and replace the denominator with the equivalent difference of squares expression.

            _____
          (√x + 1 + 2)(x - 3)
    lim                                                           eq. 2.1x12b
   x → 3     (x + 1) - 2²

When you simplify the denominator, you get

            _____
          (√x + 1 + 2)(x - 3)
    lim                                                           eq. 2.1x12c
   x → 3         x - 3

And, of course, you can cancel an x - 3 from top and bottom.

           _____
    lim   √x + 1 + 2                                              eq. 2.1x12d
   x → 3

And when you evaluate this expression at x = 3, you get 4. This may be as far as the problem asks you to go. But if it asks for a delta-epsilon proof, you will have to continue further. You then have to prove that you can evaluate it as close to x = 3 as you like, and the closer x gets to 3, the closer the expression gets to 4. To do a delta-epsilon proof of this problem, set up:

     ___________
   |√(x ± δ) + 1 + 2 - 4)|  ≤  ε                                   eq. 2.1x13a

The 4 is because we are trying to show that the limit is 4. The limit we want is as x goes to 3. So if you substitute x = 3 into the above expression, that is, for out purposes, the same as of saying |x - 3| ≤ δ .

     ___________
   |√(3 ± δ) + 1 + 2 - 4|  ≤  ε                                    eq. 2.1x13b

Simplifying further, we have

     _____
   |√4 ± δ - 2|  ≤  ε                                              eq. 2.1x13c

We now remove the absolute value brackets by breaking the problem into two parts (which is frequently the tool you have to use in order to work problems with absolute value signs in them).

Remember that we are looking for a δ > 0. When √δ > 2, then we know we are dealing with adding δ to 4 (Why? Because √x always increases with positive x):

    _____
   √4 + δ - 2  ≤  ε                                               eq. 2.1x13d

otherwise if √δ < 2 then we are subtracting δ from 4, and we get

        _____
   2 - √4 - δ  ≤  ε                                               eq. 2.1x13e

In the first case that gives us

    _____
   √4 + δ  ≤  ε + 2                                               eq. 2.1x13f

and in the second it gives

     _____
   -√4 - δ  ≤  ε - 2                                              eq. 2.1x13g

    _____
   √4 - δ  ≥  2 - ε                                               eq. 2.1x13h

Since we are looking to make epsilon closer and close to zero, we can assume that in either case, 2 - ε > 0 . Why? Because we assume we must be able to get it closer to zero than 2. If you square both sides you get

   4 + δ  ≤  4 + 4ε + ε²                                          eq. 2.1x14a

and

   4 - δ  ≥  4 - 4ε + ε²                                          eq. 2.1x14b

respectively. Now cancel the 4 in each of these and you get

   δ  ≤  4ε + ε²                                                  eq. 2.1x15a

and

   -δ  ≥  -4ε + ε²                                                eq. 2.1x15b

or equivalently

   δ  ≤  4ε - ε²                                                  eq. 2.1x15c

So which inequality should we use to establish δ, 2.1x15a or 2.1x15c? We always choose the one that gives us the smaller δ. Since ε is always positive, it should be clear that 2.1x15c is more restrictive of δ, so we choose it. More restrictive means that δ is confined to a smaller interval above zero. Remember if the δ contract satisfies a more restrictive ε expression, it will certainly also satisfy a less restrictive one.

The inequality in 2.1x15c is fine until somebody chooses an ε that is greater than or equal to 4. When that happens, it ruins our requirement that δ > 0. But δ = 3 worked when you choose ε = 1. Won't that same δ work for any ε that is greater than 1? Any δ that can bracket the function inside a smaller interval around the limit can certainly bracket it inside a larger interval around the limit. So, whenever anybody chooses ε > 1, simply choose δ = 3 instead of using 2.1x15c.

And so we have demonstrated a recipe that satisfies the contract for any ε anybody might name. And that is the objective of a delta-epsilon proof. So we're done with this problem.

Here's another limit problem that uses the difference of squares twice.

Exercises for You to Work

1) Find the limit and prove it using the delta-epsilon method of

              18x²  -  6x  -  4
     lim                       
    x → 2/3       (x - 2/3)

2) Find the limit and prove it using the delta-epsilon method of

              c(x - a) (x + b)
     lim                      
    x → a         (x - a)

3) Find the limit of

              ______
             √1 + Ax - 1
     lim                
    x → 0        x

where A is any real number. I'll give you a break and not make you prove it using the delta-epsilon method, but you won't get any objections from me if you try it anyway.

Hint: Start by applying the difference of squares method, and see if it leads to any cancellations. BTW: Your answer should have an A in it.

4) The function, ln(x), is zero at x = 1. For all other positive x, ln(x) is nonzero. At x = 0, ln(x) is undefined. What can you say about the following limits?

             A ln(x)
     lim            
    x → 1     ln(x)

and

             A ln(x)
     lim            
    x → 0     ln(x)

where A is any real number.

Yet Another Worked Problem

There is another type of problem that is likely to show up on either your homework or on and exam (or both). That is, for example, demonstrating, using the delta-epsilon method, that

    lim   x² - 5x + 8  =  2
   x → 3

Notice that there is no denominator here that goes to zero as x goes to 3. Notice also that if f(x) = x² - 5x + 8, then f(3) = 2. And you might even ask yourself, "What's the point? Why take a limit when you can already evaluate this function at x = 3?" In the next unit, which is on continuity, this will become more apparent. Continuity depends on the value of the function being the same as the value of the limit. But the other point is to give you more practice at applying delta-epsilon proofs.

So what would the delta-epsilon contract look like here? It would be that if I choose any positive ε, no matter how close to zero, that you should be able to tell me a positive δ such that

   |x² - 5x + 8 - 2|  <  ε

whenever |x - 3| < δ. Why did we subtract 2 in the first (that is the ε) inequality? Because 2 is the limit that we are trying to show. Of course we can simplify that first inequality to

   |x² - 5x + 6|  <  ε

Why did we subtract 3 in the second (that is the δ) inequality? Because we are trying to demonstrate a limit as x goes to 3.

There are several approaches you might take here. The easy one is to observe that x² - 5x + 6 factors easily into (x - 3)(x - 2). So now our contract is that for any ε demonstrate that there is a δ that causes

   |(x - 3)(x - 2)|  <  ε

whenever |x - 3| < δ.

First observe that there are two roots to (x - 3)(x - 2). And we want to make sure that the δ we choose keeps x close enough to 3 (which is one of the roots) so that x will be nowhere near the other root (at x = 2). The roots differ by 1. So if we restrict δ to being less that 0.25 in this case, that is guaranteed to keep x closer to 3 than to 2. Make a mental note of this, because we will come back to it later. And note that the choice of 0.25 was completely arbitrary. It could have been anything small enough to ensure that x will stay safely away from the other root.

Recall that our contract says, "whenever |x - 3| < δ." If δ is greater than |x - 3|, then

   |(x - 3)(x - 2)|  =  |x - 3| |x - 2|  <  δ |x - 2|

Why? Because this inequality is the same as multiplying the inequality, |x - 3| < δ, by |x - 2|, which is always positive. You are allowed to do that to inequalities. (Can you explain why |x - 2| cannot possibly be equal to zero?)

Therefore we can amend the ε inequality of our contract to read

   |(x - 3)(x - 2)|  <  δ |x - 2|  <  ε

or simply

   δ |x - 2|  <  ε

Well, |x - 2| is guaranteed to be greater than zero (it is guaranteed not to be equal to zero because previously we restricted δ to be less than 0.25). And this means we can divide the entire inequality by |x - 2| and it will still be true:

δ <

ε |x - 2|

As x gets close to 3, it is quite clear that |x - 2| gets close to 1. Hence this becomes

δ <

ε 1

So you take any δ less than ε, provided it is also less than 0.25, and you will have a δ that satisfies the contract. In other words, for any ε choose δ to be less than the minimum of either 0.25 or ε. And that's it -- you're done.

I did promise you a second method. The result here will look different than what we just arrived at above, but it will also satisfy the contract, and that is all it has to do.

If |x - 3| < δ, then x = 3 ± d, where 0 < d < δ. We can substitute this expression for x back into the ε inequality:

   |(3 ± d)² - 5(3 ± d) + 6|  < ε

Multiply this out, gather like terms (you get some cancellations), and you get

   |±d + d²|  <  ε

Clearly the + of the ± gives us the worse case, so we choose it. Since d is positive, we can drop the absolute values, subtract ε from both sides, and get:

   d² + d - ε  <  0

This is a quadratic inequality. If it were a quadratic equation, then the solutions would be (by applying the quadratic formula)

               ______
         -1 ± √1 + 4ε
   d  =              
               2

We said that d was positive (for the same reason that δ is positive). So we are forced to take the + of the ±. So the root we are interested in is

               ______
         -1 + √1 + 4ε
   d  =              
               2

As ε goes to zero, it is clear that d does also. To satisfy the quadratic inequality, then, we must choose

               ______
         -1 + √1 + 4ε
   d  <              
               2

So to satisfy the delta-epsilon contract, we must choose



   δ   <  d  <

         ______
   -1 + √1 + 4ε
               
         2

Such a δ will guarantee the contract is met. So this is also a legitimate delta-epsilon proof, even though I'd recommend you use the first method.

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