The author first shows an inequality, then he argues $2\log2$ is the smallest constant. I want to understand how to show that $2\log2$ cannot be replaced by a smaller constant. I don't know why he says "since $| \log 2 -\log(1+t)/t| \le \epsilon$ for all $x$ with $|1-x| \le \delta(\epsilon)$." To calculate $\lim_{x\to 1} \dfrac{1}{1-x}\int_x^1 \log(1+t)\dfrac{dt}{t} =\log 2$, I think we just use L'Hospital's rule. What is $|1-x| \le \delta(\epsilon)$? There is no $\delta(\epsilon)$ in the original question.
L'Hospital's rule is not the alpha and omega of limits computation!
The assertion comes from $\;\displaystyle\lim_{t\to 1}\frac{\log(1+t)}t=\log 2$, so for any $\varepsilon >0$ there exists a $\delta(\varepsilon)>0$ such that $$\biggl|\log 2-\frac{\ln(1+t)}t\biggr|<\varepsilon \quad\text{ for all $\,t$ such that }\;|t-1|<\delta.$$ In particular this is true for all $t\in [x,1]$ if $|x-1|<\delta$, so by the Mean value inequality, one has $$\Biggl|\int_x^1 \log 2-\dfrac{\log(1+t)}t\,dt\Biggr| \le\int_x^1 \Biggl|\log 2-\dfrac{\log(1+t)}t\Biggr|\,dt\le \varepsilon (1-x)$$ if $|x-1|<\delta$.