Finding the limit of the integral Let $f$ be continuous in [0, 1]. Assume $$0 < a < b$$
Prove that the following limit exists and determine what it is:
$$\lim_{\delta\rightarrow 0}\int_{\delta a}^{\delta b} \frac{f(x)}{x} dx$$
I got stuck on this question and I can't seem to figure it out. It seems to me that there's no reason for this limit to always exist. For instance, if I take
$$f(x) = 1$$
The integral of the harmonic function diverges near 0, so by Cauchy's critrion, the limit shouldn't exist.
So what am I missing? Does anyone have an idea?
 A: Hint:
By the integral MVT, there exists $\xi \in (\delta a, \delta b)$ such that $\xi \to 0$ as $\delta \to 0$ and  
$$\int_{\delta a}^{\delta b} \frac{f(x)}{x} \, dx = f(\xi)\log \frac{\delta b}{\delta a} $$
A: Do a change of variables by setting $y = \frac{x}{\delta}$. Then $y \in [a,b]$ and $dx = \delta dy$. Hence
$$
\int_{\delta a}^{\delta b } \frac{f(x) }{x} dx = \int_{a}^b \frac{f(\delta y)}{y} dy \to f(0) \int_a^b \frac{dy}{y} = f(0) \ln \frac{b}{a},
$$
where you need to use the continuity of $f$ to prove the passage to the limit.
A: You can predict that the limit is $f(0) \ln(b/a)$ by observing that for small enough $\delta$, by continuity of $f$,
$$\int_{\delta a}^{\delta b} \frac{f(x)}{x}dx \approx \int_{\delta a}^{\delta b} \frac{f(0)}{x}dx$$
To prove it, note that
$$\left| \int_{\delta a}^{\delta b} \frac{f(x)}{x} dx - f(0)\ln\frac{b}{a} \right| = \left| \int_{\delta a}^{\delta b} \frac{f(x)}{x} dx - f(0) \int_{\delta a}^{\delta b} \frac{dx}{x} \right| = \left| \int_{\delta a}^{\delta b} \frac{f(x) - f(0)}{x} dx \right| \le \int_{\delta a}^{\delta b} \left|\frac{f(x) - f(0)}{x}\right| dx$$
Let $\epsilon > 0$. Since $f$ is continuous, there is $\alpha > 0$ such that $0 < x < \alpha \implies |f(x) - f(0)| < \epsilon'$, where $$\epsilon' = \frac1{\ln \frac{b}{a}} \epsilon$$
Let $$\delta_0 = \frac{\alpha}{b}$$
Let $\delta < \delta_0$. For $x < \delta b$, we have $x < \alpha$, so $|f(x) - f(0)| < \epsilon'$. Hence, following the above series of inequalities,
$$\left| \int_{\delta a}^{\delta b} \frac{f(x)}{x} dx - f(0)\ln\frac{b}{a} \right| < \epsilon' \int_{\delta a}^{\delta b} \frac{dx}{x} = \epsilon$$
So we are done.
