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Math Help - Factoring stuff out of limits

  1. #1
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    Factoring stuff out of limits

    If I do this:

    \lim_{\alpha\rightarrow 0} \frac{sin\alpha}{\frac{2\alpha}{5}} = \lim_{\alpha\rightarrow 0} \frac{5sin\alpha}{2\alpha} = \lim_{\alpha\rightarrow 0} \frac{5}{2}\cdot \frac{sin\alpha}{\alpha}

    Am I allowed to do this?

    \frac{5}{2} \cdot \lim_{\alpha\rightarrow 0} \frac{sin\alpha}{\alpha} = \frac{5}{2} \cdot 1 = \frac{5}{2}

    Or did I do something in the very first steps incorrectly?
    Last edited by daigo; June 29th 2012 at 03:14 AM.
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  2. #2
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    Re: Factoring stuff out of limits

    Yes you can factor out constants
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  3. #3
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    Re: Factoring stuff out of limits

    Your derivation is correct. The theorem about the limit of product says that if the two limits on the right exist, then

    \lim\limits_{x \to p} & (f(x)\cdot g(x)) = \lim\limits_{x \to p} f(x) \cdot \lim\limits_{x \to p} g(x)

    Since \lim_{\alpha\to0}\frac{5}{2} and \lim_{\alpha\to0}\frac{\sin\alpha}{\alpha} exist, your application of the theorem is valid.
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  4. #4
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    Re: Factoring stuff out of limits

    Also I didn't want to create another thread since this is kind of relevant:

    If I have:

    \lim_{\alpha \rightarrow 0} \frac{\sin ^{2}\alpha }{\alpha^{2}} = \lim_{\alpha \rightarrow 0} (\frac{\sin\alpha }{\alpha})^{2}

    Can I do this?

    \lim_{\alpha \rightarrow 0} (\frac{\sin\alpha }{\alpha} \cdot \frac{\sin\alpha }{\alpha}) = (\lim_{\alpha \rightarrow 0} \frac{\sin\alpha }{\alpha}) \cdot (\lim_{\alpha \rightarrow 0} \frac{\sin\alpha }{\alpha}) = 1 \cdot 1 = 1

    But this wouldn't work with a variable, only constants?
    Last edited by daigo; June 29th 2012 at 03:26 AM.
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  5. #5
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    Re: Factoring stuff out of limits

    The theorem is applicable to the product of any two functions; one of them does not have to be a constant.
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