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I have a problem which I think i'm close to solve, but I got stuck. The problem is as follows: Let f be continuous on [0,1] and let enter image description here . Show that the following limit exists and find it: $$\lim_{\epsilon \rightarrow 0^+} \int_{\epsilon a}^{\epsilon b} \dfrac{f(x)}{x}dx$$ My attempt was as follows: by the fundamental theorem of calculus, because f(x)/x is continuous on (0,1], there's a function G(x) such that for all x in (0,1] G'(x)=f(x)/x so enter image description here I don't have any ideas how to get further from here. after checking for some private cases, my guess is that the limit is f(0)*ln(b/a). Any help would be appreciated, Thanks!

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    $\begingroup$ @Tamr I tried to replace your image with LaTeX, but I may have deleted the first one by accident. Sorry. Can you put it back? And to be honest, you should probably LaTeX-ize the rest of your math if you want to get more answers. $\endgroup$
    – JonathanZ
    Nov 30, 2018 at 21:56
  • $\begingroup$ Alright, working on that $\endgroup$
    – Tamir
    Nov 30, 2018 at 21:58

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Note that for any $\varepsilon>0$ we have $$\int_{a\varepsilon}^{b\varepsilon} \frac{f(0)}{x} dx = f(0) (\log(b\varepsilon)-\log(a\varepsilon)) = f(0) \log(b/a).$$ Since $f$ is continuous, for any $\varepsilon'>0$ there exists $\delta>0$ such that $|f(x)-f(0)|<\varepsilon'$ for all $|x| \le \delta$. Thus forall $0 <\epsilon <\delta/b$ we have $$\left| \int_{a\varepsilon}^{b\varepsilon} \frac{f(x)}{x} dx - f(0) \log(b/a) \right| \leq \int_{a\varepsilon}^{b\varepsilon} \frac{|f(x)-f(0)|}{x} dx \le \varepsilon' \log(b/a).$$ This proves that the limes is $f(0) \log(b/a)$.

Note that the limes is related to "prinicpal values", see in Wikipedia.

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