Evaluating a limit of an integral I am trying to solve the following problem.

Let $f:[0,\infty)\rightarrow\mathbb R$ be a continuous function and $b>a>0$ be real numbers.  Prove that $$ \lim_{\epsilon\rightarrow+0}\int_{a\epsilon}^{b\epsilon}\frac{f(x)}{x}dx = f(0)\log\frac{b}{a}.$$

If $f$ were differentiable I could use integration by parts, but I do not know what to do with general continuous $f$.
I would be grateful if you could give me a clue.
 A: This was not hard after all.
Let $I$ be the integral on the LHS, then we have
$$
\inf_{a\epsilon\le x\le b\epsilon}f(x)\log(b/a) \le I \le \sup_xf(x)\log(b/a).
$$
Since $f$ is continuous, the LHS and the RHS tends to $f(0)\log(b/a)$ as $\epsilon\rightarrow+0$, and thus $I\rightarrow f(0)\log(b/a)$.
A: Since $f$ is continuous it is bounded, thus by DCT:
$$\lim_{\epsilon\to 0}\int_{a\epsilon}^{b\epsilon} \frac{f(x)}{x}\,dx\overset{x\mapsto\epsilon x}=\lim_{\epsilon\to 0}\int_a^b \frac{f(\epsilon x)}{x}\,dx=\int_a^b \frac{f(0)}{x}\,dx=f(0)\ln \frac{b}{a}$$
A: $$\int_{a\epsilon}^{b\epsilon}\frac{f(x)}{x}dx = \int_{a\epsilon}^{b\epsilon}\frac{f(x)-f(0)}{x}dx + \int_{a\epsilon}^{b\epsilon}\frac{f(0)}{x}dx .$$
The second integral evaluates to the answer you want, so the first one we want to show is zero in the limit. From the mean value theorem:
$$\int_{a\epsilon}^{b\epsilon}\frac{f(x)-f(0)}{x}dx = \frac{f(\xi)-f(0)}{\xi} \varepsilon (b-a) $$
for some $a \varepsilon \le \xi \le b \varepsilon$. Therefore $$\frac{1}{b} \le \frac{\varepsilon}{\xi}\le \frac{1}{a},$$ so
$$(f(\xi)-f(0) )\frac{(b-a)}{b} \le \frac{f(\xi)-f(0)}{\xi} \varepsilon (b-a) \le (f(\xi)-f(0)) \frac{(b-a)}{a}$$
As $\varepsilon \rightarrow 0+$, so does $\xi \rightarrow 0+$, therefore by the squeeze test this is zero.
