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}
$