# $\lim_{x\rightarrow \infty }\frac{f(x)}{x}=1$ implies $\lim_{x\rightarrow \infty }f'(x)=1$

I don't know how to prove the following statement. any help please..

if $$f$$ is differentiable function on $$\mathbb{R}$$ and $$\lim_{x\rightarrow \infty }\frac{f(x)}{x}=1$$, then $$\lim_{x\rightarrow \infty }f'(x)=1$$.

• @cansomeonehelpmeout $\lim_{x\rightarrow \infty }\frac{\sin x}{x}=0$ – sera Apr 4 at 10:12

You cannot prove it, since it is not true. Take, for instance, $$f(x)=x+\sin(x^2)$$. Then you do have $$\lim_{x\to\infty}\frac{f(x)}x=1$$. However, $$f'(x)=1+2x\cos(x^2)$$, and so it is not true that $$\lim_{x\to\infty}f'(x)=1$$.
• No. It it exists, it can only be $1$. This follows from the mean value theorem. – José Carlos Santos Apr 4 at 10:21