$\displaystyle \lim_{x \to 1}\frac{1}{x-1}+\frac{1}{x^2-3x+2}$
$\displaystyle \lim_{x \to 1}\frac{1}{x-1}+\frac{1}{(x-1)(x-2)}$
This is where I get lost. Do I multiply $\displaystyle (x-1)$ to both sides?
$\displaystyle \lim_{x \to 1}\frac{1}{x-1}+\frac{1}{x^2-3x+2}$
$\displaystyle \lim_{x \to 1}\frac{1}{x-1}+\frac{1}{(x-1)(x-2)}$
This is where I get lost. Do I multiply $\displaystyle (x-1)$ to both sides?
You need a common denominator.
So multiply the first fraction by $\displaystyle \frac{x - 2}{x - 2}$.
Then it becomes:
$\displaystyle \lim_{x \to 1}\frac{x - 2}{(x - 1)(x - 2)} + \frac{1}{(x - 1)(x - 2)}$
$\displaystyle = \lim_{x \to 1}\frac{x - 2 + 1}{(x - 1)(x - 2)}$
$\displaystyle = \lim_{x \to 1}\frac{x - 1}{(x - 1)(x - 2)}$
$\displaystyle = \lim_{x \to 1}\frac{1}{x - 2}$.
You should be able to go from here.