TITLE: Show integral limit
QUESTION [10 upvotes]: How to show that:
$$I:=\lim_{\epsilon \to 0^+} \int_{\epsilon}^{2\epsilon} \frac{1}{\ln{(1+x)}}dx=\ln{2}$$
Using the mean value theorem for integrals I can show that $\frac{1}{2 }\leq I \leq 1$, but I'm not able to show that $I=\ln{2}$. Any hints?

REPLY [2 votes]: Make the substitution $x=\epsilon t$, to give
$$
I=\lim_{\epsilon\to0^+}\int_1^2 \frac\epsilon{\ln(1+\epsilon t)}\ dt = \int_1^2 \lim_{\epsilon\to0^+}\frac\epsilon{\ln(1+\epsilon t)}\ dt
$$
Note that moving the limit inside the integral can fail in some situations (even with constant limits)... fortunately, we can do it if the function converges uniformly on the interval, and this function does.
Evaluating the limit, we get
$$
I = \int_1^2 \frac1t\ dt = \big[\ln t\big]_1^2= \ln 2-\ln 1 = \ln 2
$$