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!

• is this homework you were given? – The Great Duck Mar 12 '17 at 23:06
• It's not true, if the derivative has no limit, as in the case $x +\sin x$. – Lubin Mar 12 '17 at 23:07
• @Lubin Thanks for the comment. We assume the limit of the derivative exists. – Tuyet Mar 12 '17 at 23:07
• @TheGreatDuck No I am thinking about some stuff not required by the coursework. I have been thinking about this for a while but could not work it out, probably because I have not learned techniques to solve this, so I think it is better to ask for help. – Tuyet Mar 12 '17 at 23:14
• @Lubin true but the equation would still be true in that both are of the same algebraic form and both lead to nonexistence. – The Great Duck Mar 13 '17 at 7:32

$$\frac{f(2x)}{2x}-\frac{f(x)}{x}=\frac{f(2x)-f(x)}{2x}-\frac{f(x)}{2x}$$
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...
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 $\mathop \lim\limits_{x \to +\infty } \frac{{f(x)}}{x} = \lim \limits_{x \to +\infty } f'(x)$
The proposition is true, but the hypothesis is stronger than needed. Instead of assuming that $\mathop {\lim }\limits_{x \to \infty } \frac{{f(x)}}{x}$ exists, it is enough to require that $\mathop {\lim }\limits_{x \to \infty } |f(x)| = \infty$.
Then, provided that $\mathop {\lim }\limits_{x \to \infty } f'(x)$ exists, the result follows by direct application of L'Hôpital's rule.