Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.2 Limits: A Numerical and Graphical Approach - Exercises - Page 54: 44


$\displaystyle\lim_{x\rightarrow -3^{-}} \dfrac{x^2}{x^2-9}=\infty$ $\displaystyle\lim_{x\rightarrow -3^{+}} \dfrac{x^2}{x^2-9}=-\infty$ The limit does not exist

Work Step by Step

We have to estimate the limit: $\displaystyle\lim_{x\rightarrow -3\pm} \dfrac{x^2}{x^2-9}$ Graph the function: Therefore we get: $\displaystyle\lim_{x\rightarrow -3^{-}} \dfrac{x^2}{x^2-9}=\infty$ $\displaystyle\lim_{x\rightarrow -3^{+}} \dfrac{x^2}{x^2-9}=-\infty$ As the left hand limit and the right hand limit are not equal, the limit of the function in -3 does not exist. The function tends to $\infty$ when $x\rightarrow -3^{-}$ and to $-\infty$ when $x\rightarrow -3^{+}$. There is a vertical asymptote $x=-3$.
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