Answer
$$\lim_{x\to-\infty}x\ln\Bigg(1-\frac{1}{x}\Bigg)=-1$$
Work Step by Step
$$A=\lim_{x\to-\infty}x\ln\Bigg(1-\frac{1}{x}\Bigg)$$
$$A=\lim_{x\to-\infty}\frac{x\ln\Big(1-\frac{1}{x}\Big)}{1}$$
Here we must divide both numerator and denominator by $x$, which means
$$A=\lim_{x\to-\infty}\frac{\ln\Big(1-\frac{1}{x}\Big)}{\frac{1}{x}}$$
We take $u=\frac{1}{x}$. As $x\to-\infty$, $u\to0$.
$$A=\lim_{u\to0}\frac{\ln(1-u)}{u}$$
$\lim_{u\to0}\ln(1-u)=\ln(1-0)=\ln1=0$ and $\lim_{u\to0}u=0$.
This is an indeterminate form of $\frac{0}{0}$. Applying L'Hospital's Rule, we have,
$$A=\lim_{u\to0}\frac{[\ln(1-u)]'}{u'}$$
$$A=\lim_{u\to0}\frac{\frac{1}{1-u}(1-u)'}{1}$$
$$A=\lim_{u\to0}-\frac{1}{1-u}$$
$$A=-\frac{1}{1-0}=-1$$
