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Math Help - did i solved this limit correctly

  1. #1
    MHF Contributor
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    did i solved this limit correctly

    f(x) and g(x) are differentiable on 0
    f(0)=g(0)=0
    <br />
\lim _{x->0}\frac{cos(f(x))-cos(g(x))}{x^2}=\lim _{x->0}\frac{-2sin(\frac{f(x)+g(x)}{2})sin(\frac{f(x)-g(x)}{2})}{x^2}=-2<br />

    because i can use (sin x)/x=1 here
    is it ok??
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by transgalactic View Post
    f(x) and g(x) are differentiable on 0
    f(0)=g(0)=0
    <br />
\lim _{x->0}\frac{cos(f(x))-cos(g(x))}{x^2}=\lim _{x->0}\frac{-2sin(\frac{f(x)+g(x)}{2})sin(\frac{f(x)-g(x)}{2})}{x^2}=-2<br />

    because i can use (sin x)/x=1 here
    is it ok??
    No because f(x)=g(x) is permitted by your conditions, and so if f(x) \not \equiv 0 then the limit in the middle is 0

    CB
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