Show integral limit

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 $$\int_{\epsilon} ^{2\epsilon}\frac{dt}{t}=\log 2$$

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.