Limits to Trig Functions

**EliteNewbz** · March 17th 2010, 08:56 PM

$\lim_{x \rightarrow 0} \frac{3x-sinx}{4x} $
$ \lim_{x \rightarrow \frac{\pi}{4}} \frac{sinx-cosx}{cos 2(x)}$

I'm not sure how to approach either of these.

Please and thanks.

**drumist** · March 17th 2010, 10:06 PM

$\lim_{x \to 0} \frac{3x-\sin x}{4x} ~=~ \lim_{x \to 0} \left(\frac{3x}{4x} - \frac{\sin x}{4x}\right)$

$~=~ \lim_{x \to 0} \left(\frac{3x}{4x}\right) - \lim_{x \to 0} \left(\frac{\sin x}{4x}\right) ~=~ \lim_{x \to 0} \left(\frac{3x}{4x}\right) - \frac{1}{4} \lim_{x \to 0} \left(\frac{\sin x}{x}\right)$

Can you finish this?

For the second problem, you need to know the identity:

$\cos (2x) = \cos^2 x - \sin^2 x$

Does this help?

**DeMath** · March 18th 2010, 01:12 AM

Originally Posted by EliteNewbz

$\lim_{x \rightarrow \frac{\pi}{4}} \frac{sinx-cosx}{cos 2(x)}$

I'm not sure how to approach either of these.
Please and thanks.

${\color{red}\boxed{{\color{black}\cos{2\theta}=\co s^2{\theta}-\sin^2{\theta}}}}$

$\begin{gathered}\lim\limits_{x\to\pi/4}\frac{\sin{x}-\cos{x}}{\cos{2x}}=\lim\limits_{x\to\pi/4}\frac{\sin{x}-\cos{x}}{\cos^2{x}-\sin^2{x}}=\hfill\\=-\lim\limits_{x\to\pi/4}\frac{\cos{x}-\sin{x}}{(\cos{x}-\sin{x})(\cos{x}+\sin{x})}=\hfill\\=-\lim\limits_{x\to\pi/4}\frac{1}{\cos{x}+\sin{x}}=-\frac{1}{\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}}=-\frac{\sqrt{2}}{2}.\hfill\\\end{gathered}$

**simplependulum** · March 18th 2010, 01:24 AM

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

we have

$\sin(x) - \cos(x) = \sqrt{ (1)^2 + (-1)^2 } \sin( x - \frac{\pi}{4} )$

and consider

$\cos(2x) = \sin( \frac{\pi}{2} - 2x )$

$\sin[2 ( \frac{\pi}{4} - x )]$

so the limit

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

$= \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}\sin( x - \frac{\pi}{4} )}{ - \sin[2 ( x - \frac{\pi}{4} )]}$

Now $x - \frac{\pi}{4} \to 0$

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

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