# 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}

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