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Math Help - Find the limit for the given function.

  1. #1
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    Find the limit for the given function.

    lim as x approaches 0 \frac{\sin(5x)}{\sin(6x)}
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  2. #2
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    Quote Originally Posted by superduper1 View Post
    lim as x approaches 0 \frac{\sin(5x)}{\sin(6x)}
    Utilise the following limit

    \displaystyle\lim_{x\rightarrow0}\frac{sinx}{x}=1

    \displaystyle\frac{sin5x}{sin6x}=\frac{5xsin5x}{5x  }\ \frac{6x}{6xsin6x}=\frac{5x}{6x}\ \frac{sin5x}{5x}\ \frac{6x}{sin6x}
    Last edited by Archie Meade; August 29th 2010 at 09:50 AM.
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    answer 5/6
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    Yes,

    \displaystyle\lim_{6x\rightarrow\ 0}\frac{sin(6x)}{6x}=1

    and

    \displaystyle\lim_{5x\rightarrow\ 0}\frac{sin(5x)}{5x}=1

    The limit is often evaluated using the \frac{sinx}{x} fraction,
    though it's unnecessary if we write

    \displaystyle\lim_{y\rightarrow\ 0}siny=y

    Hence, \displaystyle\lim_{x\rightarrow\ 0}\frac{sin(5x)}{sin(6x)}=\frac{\displaystyle\frac  {5x}{5x}\lim_{5x\rightarrow\ 0}sin5x}{\displaystyle\frac{6x}{6x}\lim_{6x\righta  rrow\ 0}sin6x}=\frac{5}{6}
    Last edited by Archie Meade; August 29th 2010 at 09:57 AM.
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  5. #5
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    Quote Originally Posted by superduper1 View Post
    lim as x approaches 0 \frac{\sin(5x)}{\sin(6x)}
    Since this tends to \frac{0}{0} you can use L'Hospital's Rule...


    \lim_{x \to 0}\frac{\sin{5x}}{\sin{6x}} = \lim_{x \to 0}\frac{\frac{d}{dx}(\sin{5x})}{\frac{d}{dx}(\sin{  6x})}

     = \lim_{x \to 0}\frac{5\cos{5x}}{6\cos{6x}}

     = \frac{5\cos{(5\cdot 0)}}{6\cos{(6\cdot 0)}}

     = \frac{5 \cdot 1}{6 \cdot 1}

    = \frac{5}{6}.
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