Finding the limit of the integral

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.

Hint:

By the integral MVT, there exists $\xi \in (\delta a, \delta b)$ such that $\xi \to 0$ as $\delta \to 0$ and

$$\int_{\delta a}^{\delta b} \frac{f(x)}{x} \, dx = f(\xi)\log \frac{\delta b}{\delta a} $$

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.