<s>[INST] 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 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?
 [/INST] Integration by parts is a reasonable choice. Since:
$$ \int \frac{t}{t^4+4}\,dt = \frac{1}{4}\arctan\left(\frac{t^2}{2}\right)\tag{1}$$
we have:
$$ \int_{0}^{x}\frac{t\log(1+t)}{t^4+4}=\frac{\log(1+x)}{4}\arctan\left(\frac{x^2}{2}\right)-\frac{1}{4}\int_{0}^{x}\frac{\arctan\frac{t^2}{2}}{1+t}\,dt \tag{2}$$
and by considering the Taylor series of $\log(1+x)$ and $\arctan\left(\frac{x^2}{2}\right)$ in a neighbourhood of the origin we easily get:
$$\lim_{x\to 0}\frac{1}{x^3} \int_{0}^{x}\frac{t\log(1+t)}{t^4+4}=\frac{1}{4}\cdot\frac{1}{2}-\frac{1}{4}\cdot\frac{1}{6}=\color{red}{\frac{1}{12}}\tag{3}$$
as wanted.
As pointed by Patrick Da Silva in the comments, we can also skip the integration-by-parts step, since we are just integrating $\frac{t^2}{4}+o(t^2)$ between $0$ and $x$, hence we trivially have $\frac{x^3}{12}+o(x^3)$.
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