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Math Help - Limits Problem, can someone help with this?

  1. #16
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Here something useful...

    lim_{x\to 0} \frac{sin x}{x}=1

    in your case (q. a):


    lim_{x\to 0} \frac{sin 4x}{9x}=\frac{4}{9}lim_{x\to 0} \frac{sin 4x}{4x}

    Try apply this also for b.
    How did you go from sin4x/9x to 4/9(sin4x/4x?) I don't get how you get the 4x on the bottom
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  2. #17
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Nikhiln25 View Post
    How did you go from sin4x/9x to 4/9(sin4x/4x?) I don't get how you get the 4x on the bottom


    \frac{sin4x}{9x}=\frac{sin4x}{4x\cdot \frac{9}{4}}=\frac{4}{9} \frac{sin4x}{4x}


    yes?
    Last edited by Also sprach Zarathustra; July 7th 2010 at 08:14 PM. Reason: Education editing
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