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.