Answer
$$\lim_{x\to1}\frac{x\sin(x-1)}{2x^2-x-1}=\frac{1}{3}$$
Work Step by Step
$$A=\lim_{x\to1}\frac{x\sin(x-1)}{2x^2-x-1}$$
As $x\to1$, $[x\sin(x-1)]\to[1\times\sin(1-1)]=1\times\sin0=0$ and $(2x^2-x-1)\to(2\times1^2-1-1)=2-2=0.$
Therefore, we have here an indeterminate form of $\frac{0}{0}$. L'Hospital's Rule would thus be applied:
$$A=\lim_{x\to1}\frac{[x\sin(x-1)]'}{[2x^2-x-1]'}$$
$$A=\lim_{x\to1}\frac{\sin(x-1)+x\cos(x-1)}{4x-1}$$
$$A=\frac{\sin(1-1)+1\times\cos(1-1)}{4\times1-1}$$
$$A=\frac{\sin0+1\times\cos0}{3}$$
$$A=\frac{0+1\times1}{3}$$
$$A=\frac{1}{3}$$
