## Calculus: Early Transcendentals 8th Edition

Infinite discontinuity at $-2$. See graph.
$f(-2)$ is undefined. To try and visualize this we use: $\lim\limits_{x \to -2^{-}} f(x) = \lim\limits_{x \to -2^{-}} \frac{1}{x+2}= -\infty$ $\lim\limits_{x \to -2^{+}} f(x) = \lim\limits_{x \to -2^{+}} \frac{1}{x+2}= \infty$ Both of these answers are not the same, so the $\lim\limits_{x \to -2} f(x)$ does not exist.