$\lim_{ x \to\frac\pi4 } \frac{1-\tan(x)}{\sin(x)-\cos(x)}$ without L'Hospital rule

Question

I have to compute this limit $$\lim_{ x \to\frac\pi4 } \frac{1-\tan(x)}{\sin(x)-\cos(x)}=\lim_{ x \to \frac\pi4 } -\frac{1}{ \cos x }=-\sqrt{2}$$ Is it right? In my book the suggested solution is $-\frac{1}{ \sqrt{2} }$

Answer 1

What you did is correct. The result is $-\sqrt2$, for the reason that you provided. More precisely,$$\frac{1-\tan x}{\sin x-\cos x}=\frac{\frac{\cos x-\sin x}{\cos x}}{\sin x-\cos x}=-\frac1{\cos x}.$$

Answer 2

it is $$\frac{1-\frac{\sin(x)}{\cos(x)}}{\sin(x)-\cos(x)}=-\frac{\sin(x)-\cos(x)}{\cos(x)(\sin(x)-\cos(x))}=$$

Answer 3

$$\lim_{x\to \frac{\pi }{4}} \frac{1-\tan x}{\sin x -\cos x} \times \frac{\sin x+\cos x}{\sin x +\cos x} =\lim_{x\to \frac{\pi }{4}} \frac{(1-\tan x)(\sin x+\cos x) }{-\cos 2x}$$

Now :

$$\cos 2x=\frac{1-\tan^2x}{1+\tan^2x}$$

So we have :

$$\lim_{x\to \frac{\pi }{4}} \frac{(1-\tan x)(\sin x+\cos x)(1+\tan ^2x) }{-(1-\tan x)(1+\tan x)}=-\sqrt{2}$$

Answer 4

You can use the definition of the derivative for such problems as we note $\tan \pi/4=1$ and $\sin \pi/4=\cos \pi/4=\sqrt{2}/2$ .

$$-\lim_{ x \to\pi/4 } \frac{\tan x-1}{\sin x-\cos x }=-\lim_{x\to\pi/4}\dfrac{\tan x-\tan \pi/4}{x-\pi/4}\dfrac{1}{\dfrac{\sin x -\sin(\pi/4)-[\cos x -\cos(\pi/4)]}{x-\pi/4}}$$

Now use the definition of the derivative for all the terms. The limit then becomes:

$$-\lim_{ x \to\pi/4 } \frac{\tan x-1}{\sin x-\cos x }=-\left.\dfrac{d \tan(x)}{dx}\right|_{x=\pi/4}\cdot \dfrac{1}{\left.\dfrac{d \sin x}{dx}\right|_{x=\pi/4}-\left.\dfrac{d \cos x}{dx}\right|_{x=\pi/4}}$$ $$=-(1+\tan^2 [\pi/4])\dfrac{1}{\cos \pi/4 + \sin \pi/4}=\dfrac{-2}{\sqrt{2}}=-\sqrt{2}$$