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 - 1.4 Exercises: 15

Answer

$\lim\limits_{\Delta x\to0^-}\dfrac{\dfrac{1}{x+\Delta x}-\dfrac{1}{x}}{\Delta x}=-\dfrac{1}{x^2}.$

Work Step by Step

$\lim\limits_{\Delta x\to0^-}\dfrac{\dfrac{1}{x+\Delta x}-\dfrac{1}{x}}{\Delta x}=\lim\limits_{\Delta x\to0^-}\dfrac{(x)-(x+\Delta x)}{\Delta x(x)(x+\Delta x)}$ $=\lim\limits_{\Delta x\to0^-}\dfrac{-\Delta x}{\Delta x(x)(x+\Delta x)}=\lim\limits_{\Delta x\to0^-}\dfrac{-1}{(x)(x+\Delta x)}$ $=-\dfrac{1}{x(x+0)}=-\dfrac{1}{x^2}.$
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.