Answer
\[\lim_{x\rightarrow 1}\frac{x\sin (x-1)}{2x^2-x-1}=\frac{1}{3}\]
Work Step by Step
Let \[l=\lim_{x\rightarrow 1}\frac{x\sin (x-1)}{2x^2-x-1}\]
Which is $\frac{0}{0}$ form
Using L'Hopital's rule
\[l=\lim_{x\rightarrow 1}\frac{\{x\sin (x-1)\}'}{\{2x^2-x-1\}'}\]
\[l=\lim_{x\rightarrow 1}\frac{\sin (x-1)+x\cos (x-1)}{4x-1}\]
\[\Rightarrow l=\frac{\sin (1-1)+1\cos (1-1)}{4(1)-1}\]
\[\Rightarrow l=\frac{1}{3}\]
Hence, \[\lim_{x\rightarrow 1}\frac{x\sin (x-1)}{2x^2-x-1}=\frac{1}{3}\]
