Limit of a integral For $f:[0,1]\rightarrow \mathbb{R}$ continuous, let $a >0.$
\begin{eqnarray*}
L=\lim_{\varepsilon \rightarrow 0}\int_{\varepsilon a}^{\varepsilon b} \frac{f(x)}{x} dx
\end{eqnarray*}
Show that $L=f(0)\ln(\frac{b}{a})$.The first I thought was in mean value theorem for integrals. but  I don't know how can start, if can give me some hint to start. I will be gratefull.

Let $0<a<b$. By direct calculation, for any $t>0$, $\int_{ta}^{tb}\frac{f(0)}{x}dx=f(0)\ln(b/a).$
Let $\varepsilon>0$ be given. We choose $\delta\in(0,1)$ such that $|f(x)-f(0)|<\varepsilon\cdot\frac{1}{\ln(b/a)}$
whenever $x\in[0,\delta]$. (Note that $\ln(b/a)>0$.) Let
$t_{0}=\frac{\delta}{b}.$ Let $t\in(0,t_{0})$ be arbitrary.
Note that $x\in[ta,tb]\Rightarrow x\in[0,\delta]$, so we have
estimation
\begin{eqnarray*}
 &  & \left|\int_{ta}^{tb}\frac{f(x)}{x}dx-f(0)\ln(b/a)\right|\\
 & = & \left|\int_{ta}^{tb}\frac{f(x)}{x}dx-\int_{ta}^{tb}\frac{f(0)}{x}dx\right|\\
 & \leq & \int_{ta}^{tb}\left|f(x)-f(0)\right|\cdot\frac{dx}{x}\\
 & \leq & \varepsilon\cdot\frac{1}{\ln(b/a)}\cdot\int_{ta}^{tb}\frac{dx}{x}\\
 & = & \varepsilon.
\end{eqnarray*}
This shows that $\lim_{t\rightarrow0+}\int_{ta}^{tb}\frac{f(x)}{x}dx=f(0)\ln(b/a).$
