Let $f$ be Lebesgue integrable, $f(0)=0$ and assume that $f'(0)$ exists. Show that $\frac{f(x)}{x}$ is Lebesgue integrable.
I don't know what to do. The claim makes sense to me because the only point where one can get an issue is $x=0$ but at $x$ approaches to 0, the function (which is the difference quotient at 0) approaches $f'(0)$ which exists so is finite. But I don't know how to write down a proof.
Given $\epsilon > 0$ there is a $\delta > 0$ such that for $x\in [0,\delta]$ we have a constant bound
$$\left|\frac{f(x)}{x} \right| < |f'(0)| + \epsilon$$
On $[\delta , \infty)$ we have an integrable bound
$$\left|\frac{f(x)}{x} \right| < \frac{1}{\delta}|f(x)|$$
By the comparison test for integrability, the Lebesgue integrals of $x \mapsto\frac{f(x)}{x}$ exist over both $[0,\delta]$ and $[\delta, \infty)$, and, hence, over $[0,\infty)$.