Show integral limit 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?
 A: For $0<x<1$, we can use a classical "bracketting":
$$0<x-\dfrac{x^2}{2}<\ln(1+x)<x \ \iff \  $$
$$\tag{1}\dfrac{1}{x}< \dfrac{1}{\ln(1+x)}< \dfrac{1}{x-\dfrac{x^2}{2}}$$
Remark: the last fraction can be written : $$\dfrac{1}{x-\dfrac{x^2}{2}}=\dfrac{2}{2x-x^2}=\dfrac{2}{x(2-x)}=\dfrac{1}{x}+\dfrac{1}{2-x}.$$
By integration of (1) between $\varepsilon$ and $2 \varepsilon$, using 
the ascending property of the integral:
$$\underbrace{\ln(x)|_{\varepsilon}^{2 \varepsilon}}_{= \ \ln{2 \varepsilon}-\ln{\varepsilon} \ = \ \ln(2)}<I<\underbrace{\ln(x)|_{\varepsilon}^{2 \varepsilon}}_{= \  \ln(2)}-\underbrace{\ln(2-x)|_{\varepsilon}^{2 \varepsilon}}_{\to 0^-}$$
where $I$ is the looked for integral, establishing the result.
A: Hint: If you are allowed to use that for small $x$ it holds that
$$
\log(1+x) =x + O(x^2), 
$$
try proceeding that way.
A: As the function $f(x)=\ln(1+x)$ is concave and has tangent $y=x$ at origin, if $x>0$, we have 
$$(1-\varepsilon)x \le \ln(1+x)\le x\quad\text{for any }\varepsilon \; \text{such that}\;0<\varepsilon<1,$$
We deduce that
$$\frac1x\le\frac1{\ln(1+x)}\le \frac 1{1-\varepsilon}\frac1x,$$
whence
$$\int_{\varepsilon}^{2\varepsilon}\frac{\mathrm dx}{x}=\ln 2\le\int_{\varepsilon}^{2\varepsilon}\frac{\mathrm dx}{\ln(1+x)}\le \int_{\varepsilon}^{2\varepsilon}\frac{\mathrm dx}{(1-\varepsilon)x}=\frac{\ln 2}{1-\varepsilon} $$
Apply the squeezing principle when $\varepsilon\to 0\;$ to get the limit $\ln 2$.
A: 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
$$
I = \int_1^2 \frac1t\ dt = \big[\ln t\big]_1^2= \ln 2-\ln 1 = \ln 2
$$
A: Consider the function $$f(x) = \frac{1}{\log(1+x)}-\frac{1}{x}, f(0)=\frac{1}{2}$$ then $f$ is continuous at $0$. Let $$F(x) =\int_{0}^{x}f(t)\,dt,x\geq 0$$ Then we can see by continuity of $F$ that $F(2\epsilon)-F(\epsilon)\to 0$ as $\epsilon\to 0^{+}$. Your job is now complete and desired limit is $\log 2$ because $$\int_{\epsilon} ^{2\epsilon}\frac{dt}{t}=\log 2$$
A: Lucky day, you are allowed to integrate a Taylor expansion.
$\frac{1}{ln(1+x)}=\frac1x+O(1)$
$\displaystyle{\int_{\varepsilon}^{2\varepsilon}\frac{1}{ln(1+x)}=\left[C+ln(x)+O(x)\right]_{\varepsilon}^{2\varepsilon}=(C-C)+\ln(2\varepsilon)-\ln(\varepsilon)+O(\varepsilon)=\ln(\frac{2\varepsilon}{\varepsilon})+O(\varepsilon)\to\ln(2)}$
