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Math Help - Limits to Trig Functions

  1. #1
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    Limits to Trig Functions

    \lim_{x \rightarrow 0} \frac{3x-sinx}{4x}<br />
    <br />
\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.
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  2. #2
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    \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?
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by EliteNewbz View Post
    \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}
    Last edited by DeMath; March 18th 2010 at 01:28 AM.
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  4. #4
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    <br />
L =\lim_{x \to \frac{\pi}{4}} \frac{\sin(x)-\cos(x)}{cos(2x)}<br />

    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

    <br />
\lim_{x \to \frac{\pi}{4}} \frac{\sin(x)-\cos(x)}{cos(2x)}<br />

     =  \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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