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Math Help - limit question

  1. #1
    Junior Member cinder's Avatar
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    limit question

    \begin{array}{cc}\lim\\x\rightarrow0\end{array} \frac{\sin2x}{\sin3x}

    How can I solve that without using L'H˘pital's rule (which is considered illegal in my calculus class)?
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  2. #2
    TD!
    TD! is offline
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    For small x (which we have here, since we're letting x tend to 0) you can use the first order Taylor expansion for sin(x) which is x. So for small x:

    \frac{{\sin \left( {2x} \right)}}{{\sin \left( {3x} \right)}} \approx \frac{{2x}}{{3x}} \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {2x} \right)}}{{\sin \left( {3x} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{3x}} = \frac{2}{3}

    It's also possible that you have to use the 'standard limit':

    \mathop {\lim }\limits_{\alpha  \to 0} \frac{{\sin \alpha }}{\alpha }=1

    In that case, rewrite as follows:

    \mathop {\lim }\limits_{x \to 0} \frac{{2\sin \left( {2x} \right)}}{{2x}}\frac{{3x}}{{3\sin \left( {3x} \right)}} = 2\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {2x} \right)}}{{2x}} \cdot \frac{1}{3}\mathop {\lim }\limits_{x \to 0} \frac{{3x}}{{\sin \left( {3x} \right)}} = \frac{2}{3}
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  3. #3
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    Cinder I realize that this is off topic. I looked at your LaTex code an saw that you wrote that limit in array mode. I think it is easier if you write it as:
    \lim_{n\to 0}\frac{\sin 2x}{\sin 3x}
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