Answer
\[-1\]
Work Step by Step
Let \[l=\lim_{x\rightarrow -\infty}x\;\ln\left(1-\frac{1}{x}\right)\]
\[l=\lim_{x\rightarrow -\infty}\frac{\ln\left(1-\frac{1}{x}\right)}{\frac{1}{x}}\]
Which is $\frac{0}{0}$ form
By using L' Hopital's rule
\[\Rightarrow l=\lim_{x\rightarrow -\infty}\frac{\{\ln\left(1-\frac{1}{x}\right)\}'}{(\frac{1}{x})'}\]
\[l=\lim_{x\rightarrow -\infty}\frac{\left(\frac{\frac{1}{x^2}}{1-\frac{1}{x}}\right)}{\frac{-1}{x^2}}\]
\[l=\lim_{x\rightarrow -\infty}\frac{-1}{1-\frac{1}{x}}=-1\]
Hence \[l=-1\]
Hence ,
\[\lim_{x\rightarrow -\infty}x\;\ln\left(1-\frac{1}{x}\right)=-1\]
