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Math Help - Limit problem.

  1. #1
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    Limit problem.

    Can any1 help me to find the solution for this problem


    lim a sin square x b log cosx /4
    x->0

    equals to half i.e 1/2
    Find a and b.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by arvindshetty86 View Post
    Can any1 help me to find the solution for this problem

    2
    lim a sin^ x b log cosx 1
    x->0 ------------------------- = ---
    4 2

    Find a and b.
    \lim_{x\to0}\frac{a\sin^2\left(x\right)\cdot b\log\left(\cos\left(x\right)\right)}{4}=\frac{1}{  2}

    The way you posted the question is unclear and hard to interpret. Is this what you mean?
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by arvindshetty86 View Post
    Can any1 help me to find the solution for this problem

    2
    lim a sin^ x b log cosx 1
    x->0 ------------------------- = ---
    4 2

    Find a and b.
    hi can you fix your post? try using parenthesis and stuffs.. tnx
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  4. #4
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    yes Chris that's the question.Thanks for the help.
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  5. #5
    Lord of certain Rings
    Isomorphism's Avatar
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    L'Hospitals

    Quote Originally Posted by For Arvind by Chris
    \lim_{x\to0}\frac{a\sin^2\left(x\right)\cdot b\log\left(\cos\left(x\right)\right)}{4}=\frac{1}{  2}
    Hello Arvind,

    Can you recheck the question? Something is not right here...

    Why would they keep constants like that (a and b) when they can group it and call it one constant ( like c = ab) ?

    And secondly, a simple L'Hospital will show that:

    \lim_{x\to0}\sin^2 x\cdot \log\left(\cos x\right) = 0

    So I dont think a constant ab exists

    All this leads me to believe there is a typo
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