Suppose $f(x)$ is a continuous function over $[0,1]$,and $0<a<b$ find the below limit $$\lim_{t\to 0^+}{\int_{at}^{bt}\frac{f(x)}{x}}=?$$
My work is below

$\int_{at}^{bt}\frac{f(x)}{x}=\int_{at}^{bt}f(x)d(\ln(x))$

Using the mean value theorem, and $f\in c^1[0,1]$ we have $$\exists c:a<c<b \ |f(ct)=\frac{\int_{at}^{bt}f(x)d(\ln(x))}{\ln(bt)-\ln(at)}$$ so $$\lim_{t\to 0^+}{\int_{at}^{bt}\frac{f(x)}{x}}=\lim_{t\to 0^+}{\int_{at}^{bt}f(x)d(\ln(x))}=\\
\lim_{t\to 0^+}f(ct)ln(\frac{bt}{at})=f(0)\ln(\frac ba)
$$
My questions:

1:Is my working true?

2:Is there another Idea or method to solve the problem?

Thanks in advance