# Differentiable functions and existence of limits

If a function is differentiable everywhere, does it imply that the limit at $$\pm \infty$$ is either finite or it diverges to $$\pm \infty$$?

• The function $x \mapsto 0$ is differentiable everywhere. – copper.hat Jul 19 at 16:15
• @copper.hat: The limits at $\pm\infty$ of $x\mapsto0$ (both exist and) are both finite; they are both $0.$ – Will R Jul 20 at 0:42
• Yes, I do know what I was thinking. – copper.hat Jul 20 at 1:05
• Do not I mean... – copper.hat Jul 20 at 1:06

## 3 Answers

No, as the sine function shows. It has no limit at $$\pm\infty$$.

No, it doesn't. For example, $$\sin$$ and $$\cos$$ are infinitely differentiable functions, but they have no limit at $$\pm \infty$$.

Neither. Counter-examples: $$f(x)=e^{-x^2},$$ and $$f(x)=x^2.$$

• But $e^{-x^2}$ does have a finite limit at $\pm\infty$ and $x^2$ diverges to $\infty$ at $\pm\infty$. Is this not what the OP was suggesting? – Jam Jul 19 at 16:19
• What I'm saying is that one of the functions is a counterexample to one side, and the other function is a counterexample to the other. $e^{-x^2}$ shows that a differentiable function need not have infinite limits at infinity. $x^2$ shows that a differential function need not have finite limits at infinity. – Adrian Keister Jul 19 at 16:25
• Oh, I see what you mean. I was under the impression that the OP was separately considering each limit at $\pm\infty$, as opposed to considering them together. I think it's not entirely clear from their question. – Jam Jul 19 at 16:35