## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to-\infty}x\ln\Bigg(1-\frac{1}{x}\Bigg)=-1$$
$$A=\lim_{x\to-\infty}x\ln\Bigg(1-\frac{1}{x}\Bigg)$$ $$A=\lim_{x\to-\infty}\frac{x\ln\Big(1-\frac{1}{x}\Big)}{1}$$ 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, $$A=\lim_{u\to0}\frac{[\ln(1-u)]'}{u'}$$ $$A=\lim_{u\to0}\frac{\frac{1}{1-u}(1-u)'}{1}$$ $$A=\lim_{u\to0}-\frac{1}{1-u}$$ $$A=-\frac{1}{1-0}=-1$$