How to show that $\lim\limits_{x \to \infty} f'(x) = 0$ implies $\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0$?

This is an immediate consequence of L'Hopital's rule. For example, below is said L'Hospital's rule, from Rudin's $\:$ Principles of Mathematical Analysis, $\:$ 1976. Note that it requires only that the denominator $\to\infty\:,\:$ not also the numerator. For more see the Monthly papers cited here.

REMARK $\ $ L'Hospital's rule (LHR) is essentially a form of the Mean value Theorem (MVT) repacked into a form convenient for limit calculations. One can of course "unpackage" the MVT and apply it directly without any mention of LHR. It's worth emphasizing that doing so does not really avoid L'Hopital's Rule (LHR) since it is precisely the proof of LHR, only specialized to a specfic function. Further, the proof of most special cases isn't much simpler than the proof of the general case of LHR. The raison d'être of the LHR abstraction is that it encapulates such applications of the Mean Value Theorem into a conveniently applicable form, so that one can easily reuse the proof by simply invoking the rule by name, not by value, i.e. not by repeating the whole proof ("inlining" it) every time one applies it!

Here's a proof that avoids L'Hôspital's rule (though you can certainly use it, as noted by Bill; basically, the condition on $f'(x)$ ensures that the function cannot oscillate like crazy).

Let $\epsilon\gt 0$. We want to show that there exists $N\gt 0$ such that if $x\gt N$, then $$\left|\frac{f(x)}{x}\right| \lt\epsilon.$$

We know that there exists $M\gt 0$ such that $|f'(x)|\lt \frac{\epsilon}{2}$ for all $x\gt M$. By the Mean Value Theorem, if $x\gt M$, there exists $c\in (M,x)$ such that $f(x)-f(M)=f'(c)(x-M)$, so $$|f(x)-f(M)| = |f'(c)|(x-M) \lt \left(\frac{\epsilon}{2}\right) x.$$ That is, for all $x\gt M$, $$\left|\frac{f(x)}{x} - \frac{f(M)}{x}\right| \lt\frac{\epsilon}{2}.$$

As $x\to\infty$, we know $\frac{f(M)}{x}\to 0$ (since $f(M)$ is fixed). So there exists $N\gt M$ such that if $x\gt N$, then $\left|\frac{f(M)}{x}\right|\lt \frac{\epsilon}{2}$. Thus, if $x\gt N$, then $$\left|\frac{f(x)}{x}\right| \leq \left|\frac{f(x)-f(M)}{x}\right| + \left|\frac{f(M)}{x}\right| \lt \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon,$$ which is what we needed to prove. QED

I'd start like this: for $x\geq 1$ we have $${f(x)\over x}={f(1)\over x}+{1\over x}\int_1^x f^\prime(y)\, dy.$$