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

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

## marked as duplicate by Math Lover, Cesareo, José Carlos Santos calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 3 at 1:17

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