1. ## Trigo- limit

Calculate :

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

$=\lim_{x\to\frac{\pi}{4}}\frac{(\cos x-\sin x)\sin 2x}{\cos x+\sin x}=0$

2) We have $\cos 2x=2\cos^2x-1=2\left(\cos^2x-\frac{1}{2}\right)=$

$=2\left(\cos^2x-\cos^2\frac{\pi}{4}\right)=2\left(\cos x-\cos\frac{\pi}{4}\right)\left(\cos x+\cos\frac{\pi}{4}\right)$

Then $\lim_{x\to\frac{\pi}{4}}\frac{\cos 2x}{\cos\frac{\pi}{4}-\cos x}=$

$=\lim_{x\to\frac{\pi}{4}}\frac{2\left(\cos x-\cos\frac{\pi}{4}\right)\left(\cos x+\cos\frac{\pi}{4}\right)}{\cos\frac{\pi}{4}-\cos x}=$

$=-\lim_{x\to\frac{\pi}{4}}2\left(\cos x+\cos\frac{\pi}{4}\right)=-2\sqrt{2}$

Or you can use l'Hospital.

3. Hello : ANATHOR SOLUTION

but :

and :

conclusion :

Thank you

4. Originally Posted by dhiab
Calculate :
L'Hospital's Rule would be easiest to use here...

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

As $x \to \frac{\pi}{4}$, the fraction $\to \frac{0}{0}$.

Using L'Hospital, this limit is the same as

$\lim_{x \to \frac{\pi}{4}}\left(\frac{2\cos{2x} - 4\sin{2x}\cos{2x}}{-\sin{2x}}\right)$

$= \frac{-0}{-1}$

$= 0$.

5. Originally Posted by dhiab
Calculate :
As above, the easiest way to evaluate the second is with L'Hospital.

The fraction $\to \frac{0}{0}$.

So using L'Hospital, it is the same as

$\lim_{x \to \frac{\pi}{4}}\left(\frac{-2\sin{2x}}{\sin{x}}\right)$

$= \frac{-2\sin{\frac{\pi}{2}}}{\sin{\frac{\pi}{4}}}$

$= \frac{-2}{\frac{1}{\sqrt{2}}}$

$= -2\sqrt{2}$.