$\underset{x\rightarrow\infty}{\lim}\frac{f(x)}{x}=0$ Implies $\underset{x\rightarrow\infty}{f'(x)}=0$

Question

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

Answer 1

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

Answer 2

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 $-f$.