1. ## limit

hello! who can solve that without using Lhopitals rule

thank you

2. Hint:
$\frac{1}
{{\cos (2x)}} = \frac{1}
{{\cos ^2 (x) - \sin ^2 (x)}} = \frac{1}
{{\left[ {\cos (x) - \sin (x)} \right]\left[ {\cos (x) + \sin (x)} \right]}}
$

3. Another approach :

$
\frac{x(1-\tan (x))}{\cos(2x)}=-x\cdot \frac{\tan (x
)-\tan\left(\pi/4 \right)}{x-\frac{\pi}{4}}\cdot \frac{x-\frac{\pi}{4}}{\cos(2x)-\cos(2\pi/4)}$

Now recall that $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)$ and you are done.