Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.5 Evaluating Limits Algebraically - Exercises - Page 72: 30


$$ \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{\tan x-1} =\frac{1}{\sqrt 2}$$

Work Step by Step

Given $$ \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{\tan x-1} $$ let $$ f(x) = \ \frac{\sin x-\cos x}{\tan x-1} $$ Since, we have $$ f(\frac{\pi}{4})= \ \frac{\sin \frac{\pi}{4}-\cos \frac{\pi}{4}}{\tan \frac{\pi}{4}-1} =\frac{\frac{1}{\sqrt 2}-\frac{1}{\sqrt 2}}{1-1}=0$$ So, we get \begin{aligned} L&= \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{\tan x-1}\\ &= \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{\frac{\sin x}{\cos x}-1} \\ &= \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{\frac{\sin x-\cos x}{\cos x}} \\ & = \lim _{x \rightarrow \frac{\pi}{4}}\frac{\sin x-\cos x}{\sin x-\cos x} \times \cos x \\ &= \lim _{x \rightarrow \frac{\pi}{4}} \cos x\\ &=\cos \frac{\pi}{4}\\ &=\frac{1}{\sqrt 2} \end{aligned}
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.