if $f'(x)\rightarrow L$ as $ x \rightarrow \infty$, $-\infty \leq L \leq \infty $ then $ f(x)/x \rightarrow L $ as $x \rightarrow \infty$ [duplicate]

For all $\epsilon > 0$, there exists $K > 0$ such that for all $x > K$

$$|f'(x) -L| < \frac{\epsilon}{2},$$

and, if $L \neq 0$,

$$|f'(x)| < \frac{3|L|}{2}$$

By the MVT, there exists $\xi > K$ such that ,

$$f(x)= f(K) + f'(\xi)(x-K),$$

and

$$\frac{f(x)}{x} = \frac{f(K)}{x} + f'(\xi)(1-K/x).$$

Hence,

$$\left|\frac{f(x)}{x}-L\right| \leq \left|\frac{f(K)}{x}\right| + \left|f'(\xi)\frac{K}{x}\right| + |f'(\xi) - L|\\<\left|\frac{f(K)}{x}\right| + \frac{3|L|}{2}\left|\frac{K}{x}\right| + \frac{\epsilon}{2}.$$

Then for $x$ sufficiently large,

$$\frac1{|x|} \left( |f(K)| + \frac{3}{2}|L||K| \right) < \frac{\epsilon}{2},$$

and

$$\left|\frac{f(x)}{x}-L\right| < \epsilon.$$

Therefore,

$$\lim_{x \rightarrow \infty}\frac{f(x)}{x}=L.$$

A similar argument can be used if $L = 0$.

For the second part, apply L'Hospitals rule:

$$M= \lim_{x \rightarrow \infty}f(x) = \lim_{x \rightarrow \infty}\frac{e^xf(x)}{e^x}\\=\lim_{x \rightarrow \infty}[f(x) + f'(x)] = M + L, $$

and conclude that $L = 0$.