## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 2 - Limits - 2.2 Limits: A Numerical and Graphical Approach - Exercises - Page 55: 51

#### Answer

$\displaystyle\lim_{x\rightarrow 1^{-}} f(x)=3$ $\displaystyle\lim_{x\rightarrow 1^{+}} f(x)=3$ $\displaystyle\lim_{x\rightarrow 3^{-}} f(x)=-\infty$ $\displaystyle\lim_{x\rightarrow 3^{+}} f(x)=4$ $\displaystyle\lim_{x\rightarrow 5^{-}} f(x)=2$ $\displaystyle\lim_{x\rightarrow 5^{+}} f(x)=-3$ $\displaystyle\lim_{x\rightarrow 6^{-}} f(x)=\infty$ $\displaystyle\lim_{x\rightarrow 6^{+}} f(x)=\infty$

#### Work Step by Step

We have to determine the limits: $\displaystyle\lim_{x\rightarrow 1\pm} f(x)$ $\displaystyle\lim_{x\rightarrow 3\pm} f(x)$ $\displaystyle\lim_{x\rightarrow 5\pm} f(x)$ $\displaystyle\lim_{x\rightarrow 6\pm} f(x)$ As we approach 1 from the left and the right, the value of $f(x)$ approaches 3: $\displaystyle\lim_{x\rightarrow 1^{-}} f(x)=3$ $\displaystyle\lim_{x\rightarrow 1^{+}} f(x)=3$ As we approach 3 from the left, the value of $f(x)$ approaches $-\infty$, while as we approach 3 from the right, the value of $f(x)$ approaches $4$: $\displaystyle\lim_{x\rightarrow 3^{-}} f(x)=-\infty$ $\displaystyle\lim_{x\rightarrow 3^{+}} f(x)=4$ As we approach 5 from the left, the value of $f(x)$ approaches $2$, while as we approach 5 from the right, the value of $f(x)$ approaches $-3$: $\displaystyle\lim_{x\rightarrow 5^{-}} f(x)=2$ $\displaystyle\lim_{x\rightarrow 5^{+}} f(x)=-3$ As we approach 6 from the left, the value of $f(x)$ approaches $\infty$, while as we approach 6 from the right, the value of $f(x)$ approaches $\infty$: $\displaystyle\lim_{x\rightarrow 6^{-}} f(x)=\infty$ $\displaystyle\lim_{x\rightarrow 6^{+}} f(x)=\infty$

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