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? 

Note that
$$
1-\tan x = \frac{\cos x - \sin x}{\cos x}=\frac{(\cos x - \sin x)(\cos x+ \sin x)}{\cos x(\cos x+ \sin x)}=\frac{\cos^2 x - \sin^2 x}{\cos x (\cos x+ \sin x)}=
\frac{\cos 2x}{\cos x (\cos x+ \sin x)}
$$
