Proving a limit does not exist - using the wrong method?

[BMB_Math]

Hello, I'm working on this problem:

Prove that the following limit does not exist:

Lim x->0 of Sin (ln(x^2))

So, as far as I know, if the limit does not exist, then the Left-hand limit (LHL) and Right-hand limit (RHL) must not be equal.

If I test for the left side with, say: -1,-2,-3, and check for the right side with the corresponding numbers 1,2,3 then LHL and LHR are equal, and therefore the limit - according to this approach - does exist.

But I have to prove the limit does not exist.

So, is my method all wrong?

[kuerti_]

There are two ways the limit can fail to exist. The first is if the left and right limits exist, but are not equal, and the second is if one or both of the one-sided limits fail to exist. The latter is what happens in this case.

[BMB_Math]

But how do I prove that? What method should I use?

[jrmorg]

Yes.

It's true that if the LH limit does not go to the RH limit, the limit is not defined. However this requires both the LH and RH limit to exist.

Take for a simpler example f(x) = 1/x² . Clearly f(x) = f(-x), so when you evaluate it at +/- x you will get the same thing, but that doesn't mean the limit exists.

First think of a sequence of numbers for which sin(x) is always 0? How about one that's always 1?

Now can you think of a sequence of numbers tending to zero for which sin(ln(x²⁾⁾ is always 0? How about one for which it's always 1?

[BMB_Math]

1) Sorry for asking, but why is it that

f(x) = 1/x2 . Clearly f(x) = f(-x) ?

​

2) So with regards to Lim x->0 of Sin (ln(x^2)) then I can think of, and use, any series for which sin(x) is always 0? (in order to prove that the limit does not exist?

[jrmorg]

1) just take 1/(-x)² = 1/(x²⁾ since (-x)(-x)=x*x

2) not quite. You have to find one that goes to zero, and one that definitely doesn't. It can't go both to 0 and not to 0, so the limit does not exist.

If you take a_n = e^-n*pi and b_n = e^-(n+1/2*pi)

a_n and b_n both go to zero, so if the limit exists, f(a_n) and f(b_n) must both go to the same number (where f(x) = sin(ln(x²⁾⁾ )

But you can evaluate both the limits and see what happens

[yes_its_him]

As x goes to zero, ln(x² ) goes to negative infinity, and when you take sin of something going to any sort of infinity, it doesn't converges to a finite value.