Trigo- limit

**dhiab** · August 3rd 2009, 10:32 PM

Calculate :

**red_dog** · August 3rd 2009, 10:51 PM

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.

**dhiab** · August 4th 2009, 11:28 PM

Hello : ANATHOR SOLUTION

but :

and :

conclusion :

Thank you

**Prove It** · August 4th 2009, 11:54 PM

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

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

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

$= 0$ .

**Prove It** · August 4th 2009, 11:59 PM

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

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

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

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

$= -2\sqrt{2}$ .

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