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 $f$