Evaluating $\displaystyle\lim_{x\space\to\space0} \frac{1}{x^5}\int_0^{x} \frac{t^3\ln(1-t)}{t^4 + 4}\,dt$ 
Evaluate the following limit: $$\lim_{x\space\to\space0} \frac{1}{x^5}\int_0^{x}
 \frac{t^3\ln(1-t)}{t^4 + 4}\,dt$$

Any advice on how to tackle this problem ?
 A: Using Fundamental Theorem of Calculus

$$\frac{d}{dx}\int_0^{x}
 \frac{t^3\ln(1-t)}{t^4 + 4}\,dt=
 \frac{x^3\ln(1-x)}{x^4 + 4}$$


$$\begin{align}
\lim_{x\space\to\space0} \dfrac{1}{x^5}\int_0^{x}\dfrac{t^3\ln(1-t)}{t^4 + 4}\,dt&=\lim_{x\space\to\space0} \frac{1}{5x^4}\cdot
 \frac{x^3\ln(1-x)}{x^4 + 4}\tag{1}\\
&=\lim_{x\space\to\space0}
 \frac{\ln(1-x)}{(5x)(x^4 + 4)}\\
&=\lim_{x\space\to\space0}
 -\frac{\ln(1-x)}{-x}\cdot\frac{1}{5(x^4 + 4)}\\
&=-\frac{1}{20}\\
\end{align}$$

$$\lim_{x\space\to\space0} \frac{1}{x^5}\int_0^{x}
 \frac{t^3\ln(1-t)}{t^4 + 4}\,dt=-\frac{1}{20}$$


Explanation : $(1)$ Use L'Hopital's Rule
A: $$\frac{t^3\ln(1-t)}{t^4+4}=\frac{t^3}{t^4+4}(-t-\frac{t^2}2...)=-\frac{t^4}4+o(t^4),$$
and integrating,
$$-\frac{x^5}{20}+o(x^5).$$
A: Substituting $t =  xu$, we get
$$\int_0^x dt \frac{t^3 \log{(1-t)}}{t^4+4} = x^4 \int_0^1 du \frac{u^3 \log{(1-x u)}}{4+x^4 u^4} $$
so the limit is,
$$\lim_{x\to 0} \frac1{x}\int_0^1 du \frac{u^3 \log{(1-x u)}}{4+x^4 u^4} =  \lim_{x\to 0} \frac1{x}\int_0^1 du \frac{-x u^4}{4} = -\frac1{20}$$
