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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuously differentiable function such that $\underset{x\rightarrow\infty}{\lim}\frac{f(x)}{x}=0$ and suppose $\underset{x\rightarrow\infty}{f'(x)}$ exists. Then Prove that $\underset{x\rightarrow\infty}{f'(x)}=0$

I can see that if we apply L'hoptal's theorem directly to $\frac{f(x)}{x}$ then we can get the answer. But is it possible to do so without knowing the value of $\underset{x\rightarrow\infty}{f(x)}$

On the similar problem: found here, they have given the existence of $\lim_{x\rightarrow\infty} f(x)$. But in this particular problem they haven't

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marked as duplicate by Math Lover, Cesareo, José Carlos Santos calculus Jan 3 at 1:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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As stated the result is wrong. Take $f(x)=\sin x^2$.

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  • $\begingroup$ Sorry I edited. $\endgroup$ – DD90 Jan 2 at 21:52
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Suppose $\lim_{x \to \infty} f'(x) >a >0$. Then $f(n)-f(n-1) > a $ for $n$ suffciently large, say for $n \geq n_0$ [This is by MVT]. Then $f(n) \geq f(n_0)+ a (n-n_0)$ for $n \geq n_0$ which contradicts the hypothesis that $\frac {f(x)} x\to 0$. For the case $\lim_{x \to \infty} f'(x) <0$ simply replace $f$ by $-f$.

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