Expert: Paul Klarreich - 1/9/2006

Question
Hello! We're studying the precise definition of limits in my calculus course. I think I understand how to show a limit is correct, but I'm having trouble with another homework problem. I don't understand how to prove a limit is false. The question asks: "Show that the assertion "the limit of 3x as x -->2 is 5" is false. To do this, it is necessary to exhibit a positive number e (epsilon) such that there is no response number d (delta) > 0."
  I understand about choosing a number e such that there is no response number d. But how do I know what e to choose? And then how do I prove that |f(x) - L| is not > e?
  I appreciate any help! I've scoured the internet for some kind of help, and can't find any, or maybe I'm being too dense to see it!

Answer
Hi, Kayla,

You wrote:
Hello! We're studying the precise definition of limits in my calculus course. I think I understand how to show a limit is correct, but I'm having trouble with another homework problem. I don't understand how to prove a limit is false.

>> Actually, you mean the limit is the wrong one, but it amounts to the same thing.

The question asks: "Show that the assertion "the limit of 3x as x -->2 is 5" is false. To do this, it is necessary to exhibit a positive number e (epsilon) such that there is no response number d (delta) > 0."

>> Well, of course, the answer is that  

lim 3x = 6, not 5
x->2

but that is not a good way to answer it.  However, we can certainly use that fact in the next part.  If we know the limit is 6, that means for some e, f(x) is close to 6, not close to 5.  So all we really have to do is choose some fairly small e, perhaps 0.1 here, and show that there is a neighborhood of x = 2 where the function values are 'far' from 5.

Keep in mind a key word in the definition of limit:  WHENEVER.

The definition says
 lim f(x) = L  
  x->c
if given e, there exists d such that

| f(x) - L | < e  WHENEVER  |x - c| < d
--------------------------------------
You wrote:

I understand about choosing a number e such that there is no response number d. But how do I know what e to choose? And then how do I prove that |f(x) - L| is not > e?

I appreciate any help! I've scoured the internet for some kind of help, and can't find any, or maybe I'm being too dense to see it!

No, your're not -- the reasoning is tricky, since it involves inequalities.  It goes like this: (I leave out the x-->2 to save typing)

If lim 3x = 5, that's the same as  lim(3x-5) = 0.

Suppose you had to prove the CORRECT limit, which is 6.  You would first do some working out:

|3x - 6| < e
3|x - 2| < e
|x - 2| < e/3

So you would decide that for any e, you can choose d = e/3.  Then you do the actual proof:

Given e, Let d = e/3.  

Then WHENEVER
|x - 2| < e/3
3|x - 2| < e
and, since 3 is positive, you can multiply it into the abs val.  So:
|3x - 6| < e
>> END OF PROOF OF THE CORRECT LIMIT

Now how about proving that  lim 3x = 5  is wrong?

Of course it is off by 1.  We can use that:

We are trying to show that the limit is zero, when we know it is 1.  So we want to keep the function values 'far away'.  The reasoning is: I KNOW I CAN GET WITHIN 0.5 OF 6, AND IF I DO THAT, I AM AT LEAST 0.5 AWAY.   So let's choose an e=0.3  (because it is nicely divisible by 3, that's why.)

We know that:

|3x - 6| < 0.3  whenever  |x - 2| < 0.1, because this is the right limit.

Then, if |x - 2| < 0.1, we can remove the absval:

-0.1 < x - 2 < 0.1   and multiply:
-0.3 < 3x - 6 < 0.3  and add 1 to get the 3x-5 in the middle. THIS IS THE KEY STEP.
0.7 < 3x - 5 < 1.3

So 3x - 5 is between 0.7 and 1.3, so its absval would have to be at least the smaller of those, thus:

|3x - 5| > 0.7 --  yes, greater than.

So we can now state that WHENEVER |x - 2| < 0.1, |3x - 5| > 0.7

Now this is saying that as long as x is reasonably (within 0.1) close to 2, 3x will be at least 0.7 away from 5.  

Keep in mind that the tricky part isn't that 3x is always far away from 5, obviously if x = 5/3, it is equal to it.

I hope this helps.

Paul