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

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  • $\begingroup$ The function $x \mapsto 0$ is differentiable everywhere. $\endgroup$ – copper.hat Jul 19 at 16:15
  • $\begingroup$ @copper.hat: The limits at $\pm\infty$ of $x\mapsto0$ (both exist and) are both finite; they are both $0.$ $\endgroup$ – Will R Jul 20 at 0:42
  • $\begingroup$ Yes, I do know what I was thinking. $\endgroup$ – copper.hat Jul 20 at 1:05
  • $\begingroup$ Do not I mean... $\endgroup$ – copper.hat Jul 20 at 1:06
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No, as the sine function shows. It has no limit at $\pm\infty$.

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

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

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    $\begingroup$ 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? $\endgroup$ – Jam Jul 19 at 16:19
  • $\begingroup$ 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. $\endgroup$ – Adrian Keister Jul 19 at 16:25
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    $\begingroup$ 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. $\endgroup$ – Jam Jul 19 at 16:35

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