How to solve this integral with limits? How to show that $$\lim_{\varepsilon\to0}\int_{\varepsilon a}^{\varepsilon b}\frac{f(x)}{x}dx=f(0)\ln(b/a)$$
where $a>0$ and $f:[0,1]\rightarrow\mathbb{R}$ a continuous function.
I tried integration by parts, but it doesn't work. Thanks!
 A: This is more of an intuition pump than a proof, so sorry if this isn't helpful.
$$(1) \quad \lim_{\epsilon \to 0} \ \int_{a \cdot \epsilon}^{b \cdot \epsilon} \cfrac{f(x)}{x} \ dx$$
We are given that $f(x)$ is continuous. Therefore,
$$(2) \quad \lim_{\epsilon \to {0}} \ f(a \cdot \epsilon)=\lim_{\epsilon \to {0}} f(b \cdot \epsilon)=f(0)=\lambda$$
Therefore since, $f(x)$ is constant over the interval of integration in $(1)$, we can treat it as a constant $\lambda$ and take the limit, and get,
$$(3) \quad \lim_{\epsilon \to 0} \ \int_{a \cdot \epsilon}^{b \cdot \epsilon} \cfrac{\lambda}{x} \ dx=\lambda \cdot \ln\left(\cfrac{b}{a}\right)$$
Therefore,
$$(4) \quad \lim_{\epsilon \to 0} \ \int_{a \cdot \epsilon}^{b \cdot \epsilon} \cfrac{f(x)}{x} \ dx=f(0) \cdot \ln\left(\cfrac{b}{a}\right)$$
A: HINT:
If $f$ is continuous at $0$, then for any $\nu>0$ there exists a number $\eta$ such that whenever $|x|<\eta$, $|f(x)-f(0)|<\nu$.  Now, analyze the integral
$$\left|\int_{a\epsilon}^{b\epsilon}\frac{f(x)-f(0)}{x}\, dz\right|\le\int_{a\epsilon}^{b\epsilon}\frac{|f(x)-f(0)|}{|x|}\,dx \tag 1$$
and prove that given any $\nu>0$, we can find a value of $\epsilon$ such that the right-hand side of $(1)$ is less that $\nu$.
A: Thanks for the help! I think that now we can do it.
Hence $f$ is a continuous function in [0,1], particularly in zero, for any $\eta>0$ there exists $\delta>0$ such that whenever $|x-0|<\delta$ then $|f(x)-f(0)|<\frac{\eta}{\ln(b/a)}$.
Therefore, for $|x|<\delta$
$$
\bigg| \int_{\varepsilon a}^{\varepsilon b}\frac{f(x)}{x}dx - f(0)\ln(b/a)\bigg|=\bigg| \int_{\varepsilon a}^{\varepsilon b}\frac{f(x)}{x}dx - \int_{\varepsilon a}^{\varepsilon b}\frac{f(0)}{x}dx\bigg|\leq \int_{\varepsilon a}^{\varepsilon b}\frac{|f(x)-f(0)|}{x}dx<\eta.
$$
Therefore, we have the statement.
