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Math Help - Need help solving a trig limit

  1. #1
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    Question Need help solving a trig limit

    Evaluate the following limit without the use of L'Hopital's rule:
     \lim_{x\to0}\frac{sin^2x}{\sqrt{1+xsinx}-cosx}

    My first instinct was to multiply by the conjugate, leaving me with:
     \lim_{x\to0}\frac{sin^2x(\sqrt{1+xsinx}+cosx)}{1+x  sinx-cos^2x}

    And from here I tried rearranging as best I could but could not end up with any form without zero in the denominator.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by xxlvh View Post
    Evaluate the following limit without the use of L'Hopital's rule:
     \lim_{x\to0}\frac{sin^2x}{\sqrt{1+xsinx}-cosx}

    My first instinct was to multiply by the conjugate, leaving me with:
     \lim_{x\to0}\frac{sin^2x(\sqrt{1+xsinx}+cosx)}{1+x  sinx-cos^2x}

    And from here I tried rearranging as best I could but could not end up with any form without zero in the denominator.
    \lim_{x\to0}\frac{sin^2x}{\sqrt{1+xsinx}-cosx}

    = \lim_{x\to0}\frac{sin^2x(\sqrt{1+xsinx}+cosx)}{1+x  sinx-cos^2x}

    = \lim_{x\to0}\frac{sin^2x(\sqrt{1+xsinx}+cosx)}{sin  ^2x+xsinx}

    = \lim_{x\to0}\frac{\sqrt{1+xsinx}+cosx}{1+\frac{x}{  sinx}}

    = \frac{\sqrt{1+0}+1}{1+1}

    = \frac{2}{2}

    = 1
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