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Math Help - A limit to be solved using L'Hopital's rule

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    Junior Member gusztav's Avatar
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    A limit to be solved using L'Hopital's rule

    Hi,

    I would really appreciate any help with finding the following limit:

    \lim _{x \to 0} \frac{\text{arcsin} \sqrt{\text{sin }x}}{\sqrt{2x-x^2}}

    As x \to 0, both numerator and denominator \to 0, so we can apply the L'Hopital's rule:

    =\lim _{x \to 0} \frac{\frac{cos x}{\sqrt{sin x-sin^2 x}}}{\frac{2-2x}{\sqrt{2x-x^2}}}

    But this is stil an indeterminate form ( \frac{\infty}{\infty}), and another application of the L'H rule further complicates the matter...

    Thanks for any help!
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  2. #2
    MHF Contributor
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    Quote Originally Posted by gusztav View Post
    Hi,

    I would really appreciate any help with finding the following limit:

    \lim _{x \to 0} \frac{\text{arcsin} \sqrt{\text{sin }x}}{\sqrt{2x-x^2}}

    As x \to 0, both numerator and denominator \to 0, so we can apply the L'Hopital's rule:

    =\lim _{x \to 0} \frac{\frac{cos x}{\sqrt{sin x-sin^2 x}}}{\frac{2-2x}{\sqrt{2x-x^2}}}

    But this is stil an indeterminate form ( \frac{\infty}{\infty}), and another application of the L'H rule further complicates the matter...

    Thanks for any help!
    Let's look at this a little further

    =\lim _{x \to 0} \frac{\frac{cos x}{\sqrt{sin x-sin^2 x}}}{\frac{2-2x}{\sqrt{2x-x^2}}}

    or

    = \lim _{x \to 0} \frac{1}{\sqrt{1-\sin x}} \frac{\cos x}{\sqrt{\sin x}} \cdot \frac{\sqrt{2x-x^2}}{2-2x}
    = \lim _{x \to 0} \frac{\cos x}{\sqrt{1-\sin x}(2-2x) } \cdot \frac{\sqrt{2x-x^2}}{\sqrt{\sin x} }
    = \frac{1}{2} \lim_{x \to 0} \frac{\sqrt{2x-x^2}}{\sqrt{\sin x} }
     = \frac{1}{2} \sqrt{ \lim_{x \to 0} \frac{2x-x^2}{\sin x } }
     = \frac{\sqrt{2}}{2}
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