TITLE: Find $\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t\ln (1+t)}{t^4+4}\mathrm{d}t $ without using L'hopital's rule
QUESTION [8 upvotes]: Find $$\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t\ln (1+t)}{t^4+4}\mathrm{d}t $$ without using L'hopital's rule. Using the rule, I got it as 1/12. What's a way to do it without L'hopital?

REPLY [2 votes]: $$\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t^2}{t^2+4}\cdot \underbrace{\frac{\ln (1+t)}{t}}_{\text{approaches 1 when $x\to0$}}\cdot \underbrace{\frac{t^2+4}{t^4+4}}_{\text{approaches 1 when $x\to0$}}\mathrm{d}t$$
$$=\lim_{x\to 0} \frac{x-2 \tan ^{-1}\left(x/2\right)}{x^3}=\lim_{x\to0} 1/12 - x^2/80+\mathcal{O}(x^4)=1/12.$$