Single Variable Calculus, Chapter 2, Review Exercises, Section Review Exercises, Problem 16

Determine $\displaystyle \lim \limits_{x \to 1} \left( \frac{1}{x - 1} + \frac{1}{x^2 - 3x + 2} \right)$


$\begin{equation} \begin{aligned} \lim \limits_{x \to 1} \left( \frac{1}{x - 1} + \frac{1}{x^2 - 3x + 2} \right) &= \lim \limits_{x \to 1} \left[ \frac{1}{x - 1} + \frac{1}{(x - 1)(x - 2)} \right] && \text{Factor the denominator}\\ \\ & = \lim \limits_{x \to 1} \left[ \frac{x - 2 + 1}{(x - 1)(x - 2)} \right] = \lim \limits_{x \to 1} \left[ \frac{x - 1}{(x - 1)(x - 2)} \right] && \text{Get the LCD and combine like terms}\\ \\ &= \lim \limits_{x \to 1} \frac{\cancel{x - 1}}{\cancel{(x - 1)}(x - 2)} = \lim \limits_{x \to 1} \frac{1}{x - 2} && \text{Cancel out like terms}\\ \\ &= \frac{1}{1 - 2} = \frac{1}{-1} && \text{Substitute value of $x$}\\ \\ & \fbox{$ = -1$} \end{aligned} \end{equation}$