I thought of a scenario where: if $ \lim_{x-> \infty}f(x)$ exists, which I can only imagine is a constant or asymptotic function, then it also seems like the derivative f' must also exist, but I don't know how to prove this either way. I guess if there is just a counterexample that would be easier than proving it. Is there an asymptotic function that somehow has a divergent derivative?
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Are you asking if its derivative exists everywhere? – Taliant Nov 21 '18 at 03:00
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No, at infinity. – user608672 Nov 21 '18 at 03:01
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Do you mean $\lim_{x\to\infty} f'(x)$? – Taliant Nov 21 '18 at 03:02
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I additionally mean that, yes, but also simultaneously the first case. – user608672 Nov 21 '18 at 03:03
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@JohnNash Not sure about the duplicate. In this question, $f$ is not said to be differentiable. the question is about differentiability itself (whatever the OP means by "differentiable at infinity"), not the value of the derivative. – Tom-Tom Nov 21 '18 at 08:49
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No, take for example $f(x)=\frac{\sin{x^2}}{x}$. Then $\lim_{x\to\infty} f(x) = 0$, but $f'(x) = 2 \cos{x^2} - \frac{\sin{x^2}}{x^2}$ and so $\lim_{x\to\infty} f'(x)$ does not exist.
Axe
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