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Thread: 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:
    $\displaystyle \lim_{x\to0}\frac{sin^2x}{\sqrt{1+xsinx}-cosx} $

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

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

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

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

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

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

    = $\displaystyle \frac{2}{2}$

    = $\displaystyle 1$
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