Suppose that $\lim_{x\to \infty} f'(x) = a$. Is it true that $\lim_{x\to \infty} {f(x)\over x} = a$ Suppose that $\lim_{x\to \infty} f'(x) = a$. Is it true that $\lim_{x\to \infty} {f(x)\over x} = a$
If so, can you prove it? Thanks!
 A: In case L'Hopital's Rule is not allowed :
Let $\epsilon > 0$, Then there exist larg enough $M >0 $ such that $ a-\epsilon \leq f'(x) \leq a+ \epsilon $, for all $   M\leq x $. Now Apply mean value theorem for $f$ on $[M , x ]$ then there exist $y_x \in [M, x]$ such that 
$$ a- \epsilon \leq f'(y_x) = \frac{f(x) -  f(M)}{ x - M}  \leq a+ \epsilon $$
Now letting $x \to + \infty$ we get $$ a - \epsilon \leq \limsup_{x \to + \infty} \frac{f(x)}{ x} \leq a+ \epsilon $$ for all $\epsilon > 0$  which implies $ \limsup_{x \to + \infty} \frac{f(x)}{ x} =a$, similarly $ \liminf_{x \to + \infty} \frac{f(x)}{ x} = a$ 
A: Let's assume $f'(x)\rightarrow0$ as $x\rightarrow\infty$. Then, there's a $x_\epsilon$ so that $|f'(x)|<\epsilon$ for $x>x_\epsilon.$ We can estimate
$$|f(x)|\le|f(x_{\epsilon/2})|+|f(x)-f(x_{\epsilon/2})|=|f(x_{\epsilon/2})|+|(x-x_{\epsilon/2})f'(\xi)|$$ with some $\xi\in[x_{\epsilon/2},x]$ (mean value theorem),  so $|f'(\xi)|<\epsilon/2.$ The last summand is $<x\,\epsilon/2$, so $$|f(x)/x|<\frac{|f(x_{\epsilon/2})|}{x}+\epsilon/2<\epsilon$$ for $x>2\,\frac{|f(x_{\epsilon/2})|}{\epsilon}.$
For the general case, consider $f(x)-ax$ instead of $f(x).$
To explain the downvotes: the original answer included unnecessarily heavy machinery, the reasoning was only justified with Lebesgue integrals and absolute continuity, being overkill for such a relatively simple statement. 
A: HINT
Let assume $a \gt 0$. Then, because $\lim_{x\to \infty} f'(x) = a \gt 0$ there is $M \gt 0$ such that $f'(x) \gt 0$ for all $x \ge M$. Therefore $f$ is strictly increasing on $[M, +\infty)$ and unbounded (otherwise $\lim_{x\to \infty} f'(x) =0$). It follows that $\lim_{x\to \infty} f(x) = +\infty$ and we can apply L'Hospital to $\lim_{x\to \infty} {f(x)\over x}$
Now, let's assume $a=0$. Let $g(x) = f(x) + x$. Then $\lim_{x\to \infty} g'(x)=1$ therefore $\lim_{x\to \infty} {g(x)\over x}=1$ and from here $\lim_{x\to \infty} {f(x)\over x}=0$
