Limit with trigonometric function 
Find $$\lim_{x \to \pi/4}\frac{1-\tan(x)}{\cos(2x)}$$
  without L'Hôpital.       

$$\lim_{x \to \pi/4}\frac{1-\tan(x)}{\cos(2x)}=\lim_{x \to \pi/4}\frac{\cos^2(x)+\sin^2(x)-\frac{\sin(x)}{\cos(x)}}{\cos^2(x)-\sin^2(x)}$$
How can I continue? 
 A: $\lim\limits_{x \to \frac{\pi}{4}}\frac{(1-tanx)}{cos2x}= 
\lim\limits_{x \to \frac{\pi}{4}}\frac{(cosx-sinx)}{cosx(cos^2x-sin^2x)}=
\lim\limits_{x \to \frac{\pi}{4}}\frac{(cosx-sinx)}{cosx(cosx-sinx)(cosx+sinx)}=
\lim\limits_{x \to \frac{\pi}{4}}\frac{1}{cosx(cosx+sinx)}=\frac{1}{\frac{1}{\sqrt{2}}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})}=1$
A: Notice, $$\lim_{x\to \pi/4}\frac{1-\tan x}{\cos 2x}$$
$$=\lim_{x\to \pi/4}\frac{1-\tan x}{\frac{1-\tan^2 x}{1+\tan^2 x}}$$
$$=\lim_{x\to \pi/4}\frac{(1-\tan x)(1+\tan^2 x)}{1-\tan^2 x}$$
$$=\lim_{x\to \pi/4}\frac{(1-\tan x)(1+\tan^2 x)}{(1-\tan x)(1+\tan x)}$$
$$=\lim_{x\to \pi/4}\frac{1+\tan^2 x}{1+\tan x}$$
$$=\frac{1+\tan^2 \frac{\pi}{4}}{1+\tan \frac{\pi}{4}}$$
$$=\frac{1+1}{1+1}=\color{red}{1}$$
A: Note that
$$
1-\tan x = \frac{\cos x - \sin x}{\cos x}=\frac{(\cos x - \sin x)(\cos x+ \sin x)}{\cos x(\cos x+ \sin x)}=\frac{\cos^2 x - \sin^2 x}{\cos x (\cos x+ \sin x)}=
\frac{\cos 2x}{\cos x (\cos x+ \sin x)}
$$
A: $$\lim_{x \to \frac{\pi}{4}}\frac{1-\tan(x)}{\cos(2x)}=
\lim_{x \to \frac{\pi}{4}}\frac1{\cos(x)}\lim_{x \to \frac{\pi}{4}}\frac{\cos(x)-\sin(x)}{\cos^2(x)-\sin^2(x)}=\frac1{\cos(\frac\pi4)}
\frac1{\cos(\frac\pi4)+\sin(\frac\pi4)}=1.
$$
A: hints: use these trig. identities $$\cos 2x=\cos^2x-\sin^2x=\left(\cos x+\sin x\right)\left(\cos x-\sin x\right)$$
& $$1-\tan x=\frac{\cos x-\sin x}{\cos x} $$
