Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.5 Exercises - Page 88: 7

Answer

$\lim\limits_{x\to4^-}f(x)=+\infty.$ $\lim\limits_{x\to4^+}f(x)=+\infty.$

Work Step by Step

$\lim\limits_{x\to4^-}f(x)=\lim\limits_{x\to4^-}\dfrac{1}{(x-4)^2}=\dfrac{1}{(4^--4)^2}=\dfrac{1}{(0^-)^2}=\dfrac{1}{0^+}=+\infty.$ $\lim\limits_{x\to4^+}f(x)=\lim\limits_{x\to4^+}\dfrac{1}{(x-4)^2}=\dfrac{1}{(4^+-4)^2}=\dfrac{1}{(0^+)^2}=+\infty.$
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