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Math Help - Difficult Trig Limit

  1. #1
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    Difficult Trig Limit

    I seem to be having a tough time with this one limit problem:
    Determine the limit of: \lim_{x \rightarrow 0} \frac{tanx-sinx}{x^3}
    I've tried doing multiple simplifications but I have the problem of constantly getting a \lim_{x \rightarrow 0} \frac{1-cosx}{x}=0 which can't be right, as wolfram-alpha and the answer book is telling me that the answer is \frac{1}{2}, which won't come from the zero multiplication!

    Don't suppose I could snag a push in the right direction? thanks!
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  2. #2
    MHF Contributor Calculus26's Avatar
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    If you can know L'Hopital's rule


    lim [tan(x)-sin(x)]/x^3 = lim sin(x)/x lim (sec(x)-1)/x^2

    = 1 lim[sec(x)tan(x)/2x

    = lim [sec^2(x)tan(x) +sec^3(x)]/2

    = 1/2
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by NDHS View Post
    I seem to be having a tough time with this one limit problem:
    I've tried doing multiple simplifications but I have the problem of constantly getting a \lim_{x \rightarrow 0} \frac{1-cosx}{x}=0 which can't be right, as wolfram-alpha and the answer book is telling me that the answer is \frac{1}{2}, which won't come from the zero multiplication!

    Don't suppose I could snag a push in the right direction? thanks!
    Note that \lim_{x\to 0}\frac{\tan x-\sin x}{x^3}=\lim_{x\to 0}\frac{\sec x-1}{x^2}

    Now by L'H˘pital's rule, we have \lim_{x\to 0}\frac{\sec x\tan x}{2x}\rightarrow\frac{0}{0}.

    Applying the rule one more time yields \lim_{x\to 0}\frac{\sec^3x+\sec^2x\tan x}{2}=\frac{1}{2}.

    Does this make sense?
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