Let $f:[0,1]\to\mathbb{R}$ be a continuous function. Calculate $\lim\limits_{c\to 0^+} \int_{ca}^{cb}\frac{f(x)}{x}\,dx$ I'm trying to integrate by parts but it has not been very effective.
 A: Hint Substitute $$u := \frac{x}{c}, \quad du = \frac{dx}{c}.$$
A: we have
$$g(c) = \int_{ca}^{cb} {\frac{{f(x)}}{x}dx} $$
now use the following substitution
$$\eqalign{
  & u = \frac{x}{c}  \cr 
  & du = \frac{1}{c}dx \cr} $$
to get
$$\eqalign{
  & g(c) = \int_{ca}^{cb} {\frac{{f(x)}}{x}dx}  = \int_a^b {\frac{{f(cu)}}{{cu}}cdu}   \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\, = \int_a^b {\frac{{f(cu)}}{u}du}  = \left. {f(cu)\ln (u)} \right|_a^b - c\int_a^b {\ln (u)f'(cu)du}   \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\, = \left( {f(cb)\ln (b) - f(ca)\ln (a)} \right) - c\int_a^b {\ln (u)f'(cu)du}  \cr} $$
Where we assumed that $f$ is continuous differentiable. Now letting $c$ goes to zero and using the continuity of $f$ and ${f'}$ at zero we have
$$\mathop {\lim }\limits_{c \to 0} g(c) = f(0)\ln (\frac{b}{a})$$
A: We suppose $0 < a < b$.
For $\epsilon > 0$, you can find $\delta > 0$ such that for $\vert x \vert \le \delta$ you have $\vert f(x)-f(0) \vert \le \epsilon$ (by continuity of $f$ at $0$).
Then for $\vert c \vert \max(a,b) \le \delta$, we have $[ca,cb] \subset [0,\delta]$. Hence:
$$\left\vert \int_{ca}^{cb} \frac{f(x)}{x} dx - \int_{ca}^{cb} \frac{f(0)}{x} dx \right\vert\le \int_{ca}^{cb} \frac{\vert f(x)-f(0) \vert}{x} dx \le \epsilon \int_{ca}^{cb} \frac{dx}{x}=\epsilon \ln(\frac{b}{a})$$
The RHS goes to $0$ as $\epsilon$ goes to $0$. So $$\lim\limits_{c \to 0^+} \int_{ca}^{cb} \frac{f(x)}{x} dx =\int_{ca}^{cb} \frac{f(0)}{x} dx=\color{red}{f(0)\ln(\frac{b}{a})}$$
This proof is only assuming that $f$ is integrable on $[0,1]$ and continuous at $0$.
A: Hint: Note that $f$ is continuous at $0$, so that we can say that over the interval of interest, $f$ satisfies
$$
\frac{f(0) - \epsilon}{x} \leq \frac{f(x)}{x} \leq \frac{f(0) + \epsilon}{x}
$$
when $c$ is sufficiently small.
