Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 1 - Limits and Continuity - 1.2 Computing Limits - Exercises Set 1.2 - Page 70: 41

Answer

a. The calculation is incorrect as one cannot subtract infinite values and reach $0$. b. Because the limit approaches $0$ from the right, the values of $x$ are small positive fractions which are less than $1$. Thus, $|\frac{1}{x^2}| > |\frac{1}{x}|$ with these values of $x$ as $|\frac{1}{x^2}|$ is dividing by a smaller number and hence resulting in a larger number. Thus, $\lim_{x \to 0^+} (\frac{1}{x}-\frac{1}{x^2})= -\infty$ as we are subtracting a larger number.

Work Step by Step

a. The calculation is incorrect as one cannot subtract infinite values and reach $0$. b. Because the limit approaches $0$ from the right, the values of $x$ are small positive fractions which are less than $1$. Thus, $|\frac{1}{x^2}| > |\frac{1}{x}|$ with these values of $x$ as $|\frac{1}{x^2}|$ is dividing by a smaller number and hence resulting in a larger number. Thus, $\lim_{x \to 0^+} (\frac{1}{x}-\frac{1}{x^2})= -\infty$ as we are subtracting a larger number.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.