Limit of the derivative equals zero implies the limit of the function exists

I've been tackling a few analysis problems today and have once again hit a bit of a block.

I need to either prove or find a counterexample to the following two propostions:

exists.

exists.

Noting here that I am saying a limit exists if it converges in In other words, if then it **DOES exist.** I know some take an infinite limit as not existing but I am taking it as existing for the purpose of this problem. A non-existent limit would be one like sin(x) where it just sort of oscillates or where the left limit and right limit are not equal.

So, any pointers here? I am pretty sure that both are true because I've had no luck finding a counter example. In a similar problem to this I had earlier (which was effectively the reverse) I was able to employ the Mean Value Theorem.

But basically I am completely stuck.

I thought about saying the following:

We know that if a function is differentiable then it must be continuous. We know that* a function is continuous if and only if it has a limit at every point*. Hence, if as then f(x) is continuous as

Thus f(x) has a limit as x

But I'm pretty sure I'm mis-remembering something with that claim in italics and thus not making any sense at all. Any pointers would be greatly appreciated.

Re: Limit of the derivative equals zero implies the limit of the function exists

Saying that as is not the same as saying that is differentiable. It is entirely possible that is not differentiable (and possibly not continuous) for some values of . Suppose that for every open interval , is not continuous for some . Does that negate the possibility of the limit existing for the derivative? I think it implies that there does need to be an open interval about infinity where is differentiable (and therefore continuous) as you stated. Just using definitions, the first statement is:

So, that statement is true if and only if there exists such that for all , there exists such that if then .

Consider what that statement implies. Consider what the derivative implies. Consider contrapositives. Try to figure out exactly what these statements mean before you attempt to prove or disprove anything. I believe if a counterexample exists, it will be to the second statement, and I think the first statement is probably true. A counterexample to the second will probably be a function whose limit is infinity. When you divide by infinity, you may not be able to get any solution.

Re: Limit of the derivative equals zero implies the limit of the function exists