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!


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.
