If the derivative tends to $0$ at infinity, then so does the function divided by $x$

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Suppose that $$f\colon (a,\infty) \to \mathbb{R}$$ is differentiable, and that $$\lim_{x\to\infty} f'(x) =0. $$

Show that $$\lim_{x\to\infty} \frac{f(x)}{x} = 0.$$

This can be done easily with L'Hôpital's rule, but it appears as a textbook exercise, in a chapter discussing Rolle's theorem, Mean Value Theorem, etc., before L'Hôpital's rule is introduced.

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Let $x >T$. $\frac {f(x)} x=\frac {f(x)-f(T)} x +\frac {f(T)} x=\frac {x-T} x \frac {f(x)-f(T)} {x-T} +\frac {f(T)} x$By MVT this gives $\frac {f(x)} x=\frac {x-T} x f'(s) +\frac {f(T)} x$ for some $s \in (T,x)$. Can you finish?