Answer
$$\lim\limits_{x\to1}\frac{\sin(x-1)}{x^2+x-2}=\frac{1}{3}$$
Work Step by Step
$$A=\lim\limits_{x\to1}\frac{\sin(x-1)}{x^2+x-2}$$$$A=\lim\limits_{x\to1}\frac{\sin(x-1)}{(x-1)(x+2)}$$ $$A=\lim\limits_{x\to1}\frac{\sin(x-1)}{x-1}\times\lim\limits_{x\to1}\frac{1}{x+2}$$
Let $x-1=\theta$. Since $x\to 1$ is considered, $(x-1)\to0$, so $\theta\to0$ also. $$A=\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta}\times\lim\limits_{x\to1}\frac{1}{x+2}$$ $$A=1\times\frac{1}{1+2}=\frac{1}{3}$$
