Prove $\lim\limits_{x \to +\infty } \frac{{f(x)}}{x} = \lim\limits_{x \to +\infty } f'(x)$ if both limits exist Let $f:\Bbb R \to \Bbb R$ be differentiable, and $\mathop {\lim }\limits_{x \to \infty } \frac{{f(x)}}{x}$ (the slope of some asymptote) exists and the limit of the derivative $\mathop {\lim }\limits_{x \to  + \infty } f'(x)$ exists as well, then show

$$\mathop \lim\limits_{x \to +\infty } \frac{{f(x)}}{x} = \lim \limits_{x \to +\infty } f'(x)$$

or

$$\lim\limits_{x \to +\infty } \frac{{f(x)}}{x} = \lim\limits_{x \to +\infty } \lim\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$$

This is intuitively true, but I don't know how to show it and I don't know how to deal with the mixed limit. Thanks!

$$\frac{f(2x)}{2x}-\frac{f(x)}{x}=\frac{f(2x)-f(x)}{2x}-\frac{f(x)}{2x}$$
Now the mean value theorem says that $\frac{f(2x)-f(x)}{2x}=\frac{f'(c_x)}{2}$ for some $c_x\in[x,2x]$. Take the limit of both sides...
