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? 

Notice, $$\lim_{x\to \pi/4}\frac{1-\tan x}{\cos 2x}$$
$$=\lim_{x\to \pi/4}\frac{1-\tan x}{\frac{1-\tan^2 x}{1+\tan^2 x}}$$
$$=\lim_{x\to \pi/4}\frac{(1-\tan x)(1+\tan^2 x)}{1-\tan^2 x}$$
$$=\lim_{x\to \pi/4}\frac{(1-\tan x)(1+\tan^2 x)}{(1-\tan x)(1+\tan x)}$$
$$=\lim_{x\to \pi/4}\frac{1+\tan^2 x}{1+\tan x}$$
$$=\frac{1+\tan^2 \frac{\pi}{4}}{1+\tan \frac{\pi}{4}}$$
$$=\frac{1+1}{1+1}=\color{red}{1}$$
