Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Practice Exercises - Page 440: 94

Answer

$$\infty$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{{x^4}}} - \frac{1}{{{x^2}}}} \right) \cr & {\text{evaluating the limit, we get:}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{{x^4}}} - \frac{1}{{{x^2}}}} \right) = \frac{1}{{{0^4}}} - \frac{1}{{{0^2}}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{{x^4}}} - \frac{1}{{{x^2}}}} \right) = \infty - \infty \cr & {\text{simplify }}\frac{1}{{{x^4}}} - \frac{1}{{{x^2}}} \cr & = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - {x^2}}}{{{x^4}}}} \right) \cr & {\text{evaluating the limit, we get:}} \cr & = \frac{{1 - {{\left( 0 \right)}^2}}}{{{{\left( 0 \right)}^4}}} = \frac{1}{0}=\infty \cr } $$
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