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Math Help - limit problem... #2

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    limit problem... #2

    Calculate:

    lim_{x\to 0} (cosx)^{[\frac{1}{x^2}]
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  2. #2
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    \underset{x\rightarrow 0}{\lim} ~(\cos x)^{\frac{1}{x^2}}=<br />
e^{- \frac{1}{2}}=\displaystyle \frac{1}{\sqrt{e}}

    ref attachment
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    I forget the ln....

    Thank you!
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    I forget the ln....
    Also this limit can be calculated without taking the logarithm and L'Hopital:

    \lim\limits_{x\to0}\Bigl(\cos{x}\Bigl)^{\tfrac{1}{  x^2}}=\lim\limits_{x\to0}\left[\Bigl(1+\cos{x}-1\Bigl)^{\tfrac{1}{\cos{x}-1}}\right]^{\tfrac{\cos{x}-1}{x^2}}=\exp\!\left(\lim\limits_{x\to0}\dfrac{\co  s{x}-1}{x^2}\right)=

    =\exp\!\left(-2\lim\limits_{x\to0}\dfrac{1-\cos{x}}{2x^2}\right)=\exp\!\left(-2\lim\limits_{x\to0}\dfrac{\sin^2\frac{x}{2}}{x^2}  \right)=\exp\!\left(-\dfrac{1}{2}\lim\limits_{x\to0}\dfrac{\sin^2\frac{  x}{2}}{\frac{x^2}{4}}\right)=

    =\exp\!\left[-\dfrac{1}{2}\lim\limits_{x\to0}\!\left(\dfrac{\sin  \frac{x}{2}}{\frac{x}{2}}\right)^2\right]=\exp\!\left(-\dfrac{1}{2}\cdot1^2\right)=e^{-1/2}=\dfrac{1}{\sqrt{e}}.
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