How to show that $f(x) = o{(x^2)}$ as $x \to \infty$. when $f$ is differentiable on $(0, \infty)$ and $f'(x) = o(x)$ as $x\ \to \infty$ Suppose $f$ is differentiable on $(0, \infty)$ and $f'(x) = o(x)$ as 
$x \to \infty$.
How to show  that $f(x) = o{(x^2)}$ as $x  \to \infty$?
 A: Cauchy's Mean Value Theorem says that
$$
\frac{f(b)-f(a)}{b^2-a^2}=\frac{f'(c)}{2c}\text{ for some $c$ between $a$ and $b$}
$$
Since $f'(x)=o(x)$, for any $\epsilon>0$ there is an $a$ so that if $c\ge a$ then $\dfrac{f'(c)}{c}\le2\epsilon$. Cauchy's Mean Value Theorem then says that $\dfrac{f(b)-f(a)}{b^2-a^2}\le\epsilon$. This says that $f(x)=o(x^2)$ as $x\to\infty$.
A: Apply De l'Hospital's theorem in the form $$\left[ \frac{\text{anything}}{\infty}\right].$$ Then
$$
\lim_{x \to \infty} \frac{f(x)}{x^2} = \lim_{x \to \infty} \frac{f'(x)}{2x}=0
$$
by assumption.
A: Let $k>0$. 
Then $f'(x) < \frac{k}{2} x$ for all $x >a$.
By the MVT, for all $x > a$ there exists some $c_x$ so that
$$\frac{f(x)-f(a)}{x-a}=f'(c_x) < \frac{k}{2}c_x <\frac{k}{2}x \,.$$
this implies that 
$$f(x)-f(a)< \frac{k}{2}x (x-a) < \frac{k}{2}x^2 \,.$$
Since $a$ is fixed, it is easy to finish the proof. 
A: Let $\epsilon>0$. Fix $x_0>0$, such that $|f'(x)|<\epsilon x$ for $x>x_0$. It holds 
$$ |f(x)| = \left|f(x_0) + \int_{x_0}^x f'(\xi)d\xi\right|
\le|f(x_0)|+\epsilon\left|\int_{x_0}^x\xi\,d\xi\right|
=|f(x_0)|+\frac{\epsilon}{2}\left|x^2-x_0^2\right|\,.$$
Now fix $x_1>x_0$, such that $|f(x_0)|<\frac\epsilon2x^2$ for $x>x_1$. Then for all $x>x_1$ we have
$$ |f(x)| < \frac\epsilon2x^2+\frac\epsilon2\left|x^2-x_0^2\right| \le \epsilon x^2\,, $$
qed.
