#### Answer

$$\frac{1}{3}$$

#### Work Step by Step

Given $$\lim _{x\to 1}\frac{\sin \left(x-1\right)}{x^2+x-2}$$
Since
\begin{align*}
\lim _{x\to 1}\frac{\sin \left(x-1\right)}{x^2+x-2}&=\lim _{x\to 1}\frac{\sin \left(x-1\right)}{(x-1)(x+2)}\\
&=\lim _{x\to 1}\frac{1}{x+2}\lim _{x\to 1}\frac{\sin \left(x-1\right)}{(x-1)}\\
&=\lim _{x\to 1}\frac{1}{x+2}\lim _{x-1\to 0}\frac{\sin \left(x-1\right)}{(x-1)}\\
&=\frac{1}{3}(1)==\frac{1}{3}
\end{align*}