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?
 A: A possibility would be to make the change of variable $t=ux$. Then the limit you are looking for is :
$$l=\lim_{x \rightarrow 0} \int_0^1 \dfrac{u}{u^4x^4+4} \dfrac{\log(1+ux)}{x} \, du $$
Then, using the dominated convergence theorem, you get the limit inside the integral :
$$\begin{array}{rcl}
l & = & \displaystyle \int_0^1 \lim_{x \rightarrow 0} \dfrac{u}{u^4x^4+4} \dfrac{\log(1+ux)}{x} \, du\\
  & = & \displaystyle  \int_0^1 \dfrac{u^2}{4} \, du \\
  & = & \dfrac{1}{12}
\end{array}$$
Edit : a little bit more of details for the dominated convergence theorem, you can use :
$$\dfrac{u}{u^4x^4+4} \dfrac{\log(1+ux)}{x} \leq \frac{u^2}{4}$$
using the classical $\log(1+t) \leq t$.
A: 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 $\frac{x^3}{12}+o(x^3)$.
A: Let's first analyze the limit when $x \to 0^{+}$. If $0 < t < 1$ then we know that $$t - t^{2} = t(1 - t) < \frac{t}{1 + t} < \log(1 + t) < t$$ Multiplying by $t$ we get $$t^{2} - t^{3} < t\log(1 + t) < t^{2}\tag{1}$$ Also we can easily see that $$\frac{1}{4}\left(1 - \frac{t}{4}\right) < \frac{1}{t + 4} < \frac{1}{t^{4} + 4} < \frac{1}{4}\tag{2}$$ If $0 < t < 1$ then all the terms involved in the inequalities $(1)$ and $(2)$ are positive and hence we can multiply them together to get $$\left(t^{2} - t^{3}\right)\left(\frac{1}{4} - \frac{t}{16}\right) < \frac{t\log(1 + t)}{t^{4} + 4} < \frac{t^{2}}{4}$$ or $$\frac{t^{2}}{4} + o(t^{2}) < \frac{t\log(1 + t)}{t^{4} + 4} < \frac{t^{2}}{4}$$ Integrating the above between $0$ and $x$ where $0 < x < 1$ we get $$\frac{x^{3}}{12} + o(x^{3}) < \int_{0}^{x}\frac{t\log(1 + t)}{t^{4} + 4}\,dt < \frac{x^{3}}{12}$$ and dividing by $x^{3}$ we get $$\frac{1}{12} + o(1) < \frac{1}{x^{3}}\int_{0}^{x}\frac{t\log(1 + t)}{t^{4} + 4}\,dt < \frac{1}{12}$$ and thus by Squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{1}{x^{3}}\int_{0}^{x}\frac{t\log(1 + t)}{t^{4} + 4}\,dt = \frac{1}{12}$$ A similar argument can be made for the case when $x \to 0^{-}$.
We don't need to use any advanced tools like Taylor series or L'Hospital's Rule and job is done via very simple inequalities and Squeeze theorem.
A: $$\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t^2}{t^2+4}\cdot \underbrace{\frac{\ln (1+t)}{t}}_{\text{approaches 1 when $x\to0$}}\cdot \underbrace{\frac{t^2+4}{t^4+4}}_{\text{approaches 1 when $x\to0$}}\mathrm{d}t$$
$$=\lim_{x\to 0} \frac{x-2 \tan ^{-1}\left(x/2\right)}{x^3}=\lim_{x\to0} 1/12 - x^2/80+\mathcal{O}(x^4)=1/12.$$
