Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 47

Given the function $f(x) = \displaystyle \frac{x^2-1}{|x-1|}$

a.) Find $(i) \lim\limits_{x \to 1^+} f(x) \qquad (ii) \lim\limits_{x \to 1^-} f(x)$



$
\begin{equation}
\begin{aligned}
(i) \lim\limits_{x \to 1^+} f(x) & = \lim\limits_{x \to 1^+} \displaystyle\frac{x^2-1}{x-1}\\
& = \lim\limits_{x \to 1^+} \displaystyle\frac{(x+1)\cancel{(x-1)}}{\cancel{(x-1)}}\\
& = \lim\limits_{x \to 1^+} (x+1)\\
& = 1+1\\
& = 2\\

(ii) \lim\limits_{x \to 1^-} f(x) & = \lim\limits_{x \to 1^-} \displaystyle \frac{x^2-1}{-(x-1)}\\
& = \lim\limits_{x \to 1^-} \displaystyle\frac{(x+1)\cancel{(x-1)}}{-\cancel{(x-1)}}\\
& = \lim\limits_{x \to 1^-} - (x+1)\\
& = \lim\limits_{x \to 1^-} -x-1\\
& = -1 - 1\\
& = -2
\end{aligned}
\end{equation}
$


b.)Is the $\lim\limits_{x \to 1} f(x)$ exist?
$\lim\limits_{x \to 1} f(x)$ does not exist because the left and right hand limits are different


c.) Sketch the graph of $f$