If $f(x)/x$ has no limit when $x\to\infty$ then $f'(x)$ has no limit when $x\to\infty$ Let $f$ be some continuously differentiable function on $(0, +\infty) $ such that $f(x)/x$ has no limit when $x \rightarrow +\infty$.
Prove that $f'$ has no limit at $+\infty$.
The only thing I see is that the function doesn't have horizontal asymptote.
 A: Hint: Assume that $f'(x)$ has a limit, then use L'Hôpital.
Edit: To explain in (a lot) more detail, assume that $\lim_{x\to\infty}f'(x)=L$.
First, if $L>0$ then you can easily prove that $\lim_{x\to\infty} f(x)=\infty$: There must be some $M$ with $f'(x)>L/2$ when $x>M$. It follows that $f(x)-f(M)=\int_M^xf'(t)\,dt>(x-M)L/2$, therefore $f(x)\to\infty$. Now the requirements for using L'Hôpital are fulfilled, and you get $$\lim_{x\to\infty}\frac{f(x)}{x}=\lim_{x\to\infty}f'(x)=L.$$
Similarly, if $L<0$ you turn some inequalities around in the above argument to show that $\lim_{x\to\infty}f(x)=-\infty$, and use L'Hôpital the same way, with the same conclusion.
Finally, if $L=0$ then applying the above to $g(x)=f(x)+x$, you get $\lim_{x\to\infty}g'(x)=1\ne0$, so $$\lim_{x\to\infty}\frac{g(x)}{x}=1.$$ Subsitute in the definition of $g(x)$ and subtract $1$ to get $$\lim_{x\to\infty}\frac{f(x)}{x}=0.$$
PS. This is a quick proof using a bit of machinery. The proof by user22705 is probably more instructive, but then it also requires more work.
A: This is a sketch of a solution to your problem. The advantage is that you do not need to know about L'Hopitals rule, we instead appeal directly to the mean value theorem. An important ingredient in the proof of that rule is the mean value theorem. 
Assume that $\lim_{x\rightarrow\infty}f^\prime(x)=L$.Let  $\delta >0$. Then there is an $N$ such that  $|f^\prime(x)-L|<\delta$ for $x\geq N$. By the mean value theorem we have that  $\frac{f(x)-f(N)}{x-N}=f^\prime(\xi)$ some $\xi \in (N,x)$. Hence $|\frac{f(x)-f(N)}{x-N}-L|=|f^\prime(\xi)-L|<\delta$ if $x\geq N$   . Futhremore by choosing $x$ (even) larger we have $|(\frac{x-N}{x}-1)|<\delta$ and $|\frac{f(N)}{x-N}|<\delta$. Also note that  $|\frac{x-N}{x}|\leq 2$ for large $x$. It follows that 
$$|\frac{f(x)}{x}-L|=|\frac{f(x)}{x}\frac{x-N}{x-N}-L|=$$ $$|\frac{f(x)-f(N)+f(N)}{x}\frac{x-N}{x-N}-L|=|\frac{f(x)-f(N)}{x-N}\frac{x-N}{x}-L+\frac{f(N)}{x-N}\frac{x-N}{x}|$$ $$\leq |(\frac{f(x)-f(N)}{x-N}-L)\frac{x-N}{x}+L(\frac{x-N}{x}-1)+\frac{f(N)}{x-N}\frac{x-N}{x}|$$ $$\leq|(\frac{f(x)-f(N)}{x-N}-L)||\frac{x-N}{x}|+|L||(\frac{x-N}{x}-1)|+|\frac{f(N)}{x-N}\frac{x-N}{x}|\leq$$ $$ 2\delta  +|L|\delta +2\delta $$
The last expression can be made smaller than any given $\epsilon>0$, by picking a small enough $\delta$.
