Answer
$$\lim\limits_{x\to1}\Big(\frac{1}{x-1}+\frac{1}{x^2-3x+2}\Big)=-1$$
Work Step by Step
$$A=\lim\limits_{x\to1}\Big(\frac{1}{x-1}+\frac{1}{x^2-3x+2}\Big)$$
$$A=\lim\limits_{x\to1}\Big[\frac{1}{x-1}+\frac{1}{(x-1)(x-2)}\Big]$$
$$A=\lim\limits_{x\to1}\frac{(x-2)+1}{(x-1)(x-2)}$$
$$A=\lim\limits_{x\to1}\frac{x-1}{(x-1)(x-2)}$$
$$A=\lim\limits_{x\to1}\frac{1}{x-2}$$
$$A=\frac{1}{1-2}$$
$$A=-1$$
