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

Question

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 ?

Answer 1

Using Fundamental Theorem of Calculus

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

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

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

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Explanation : $(1)$ Use L'Hopital's Rule

Answer 2

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,

$$\lim_{x\to 0} \frac1{x}\int_0^1 du \frac{u^3 \log{(1-x u)}}{4+x^4 u^4} = \lim_{x\to 0} \frac1{x}\int_0^1 du \frac{-x u^4}{4} = -\frac1{20}$$

Answer 3

$$\frac{t^3\ln(1-t)}{t^4+4}=\frac{t^3}{t^4+4}(-t-\frac{t^2}2...)=-\frac{t^4}4+o(t^4),$$ and integrating, $$-\frac{x^5}{20}+o(x^5).$$