Finding $\lim\limits_{a\to 1}\int_{0}^{a}x\ln(1-x)dx$ 
Calculate
  $$\lim_{a\to 1}\int_{0}^{a}x\ln(1-x)dx, a\in (0,1)$$

I calculate the integral but when I calculate the limit I get $\ln(0)$ and the limit should be $-\frac{3}{4}$.
How to approach the exercise in other way?
 A: Here’s another approach:
$$\lim_{a\to 1}\int_{0}^{a}x\log(1-x)dx=-\lim_{a\to 1}\int_{0}^{a}\sum_{n\ge 1}{\frac{x^{n+1}}{n}}dx=-\lim_{a\to 1}\sum_{n\ge1}{\int_{0}^{a}{\frac{x^{n+1}}{n}}}dx=-\lim_{a\to 1}\sum_{n\ge 1}{\frac{a^{n+2}}{n(n+2)}}$$
$$=-\sum_{n\ge 1}{\frac{1}{n(n+2)}}=-\frac{1}{2}\sum_{n\ge 1}{\left(\frac{1}{n}-\frac{1}{n+2}\right)}= -\frac{1}{2}\sum_{n\ge 1}{\left(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2}\right)}$$
$$ =-\frac{1}{2}{\overbrace{\sum_{n\ge 1}{\left(\frac{1}{n}-\frac{1}{n+1}\right)}}}^{=1}-\underbrace{\frac{1}{2}\sum_{n\ge 1}{\left(\frac{1}{n+1}-\frac{1}{n+2}\right)}}_{\text{this is 1/2}}$$
$$=-\frac{1}{2}\times 1-\frac{1}{2}\times\frac{1}{2}=-\frac{3}{4}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\bbox[10px,#ffd]{\left.\lim_{a \to 1^{\Large -}}
\int_{0}^{a}x\ln\pars{1 - x}\,\dd x\,\right\vert_{\ a\ >\ 0}}
\\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\,&
\lim_{a \to 1^{\Large -}}\bracks{%
{1 \over 2}\,a^{2}\ln\pars{1 - a} -
\int_{0}^{a}\pars{{1 \over 2}\,x^{2}}{-1 \over 1 - x}\,\dd x}
\\[5mm] = &\
\lim_{a \to 1^{\Large -}}\bracks{%
{1 \over 2}\,a^{2}\ln\pars{1 - a} -
{1 \over 2}\int_{0}^{a}\pars{1 + x - {1 \over 1 - x}}\,\dd x}
\\[5mm] = &\
\lim_{a \to 1^{\Large -}}\bracks{%
{1 \over 2}\,a^{2}\ln\pars{1 - a} -
{1 \over 2}\,a - {1 \over 4}\,a^{2} -
{1 \over 2}\,\ln\pars{1 - a}}
\\[5mm] = &\ \bbx{-\,{3 \over 4}}
\end{align}
