# limit problem... #2

• Jul 9th 2010, 10:27 PM
Also sprach Zarathustra
limit problem... #2
Calculate:

$\displaystyle lim_{x\to 0} (cosx)^{[\frac{1}{x^2}]$
• Jul 9th 2010, 11:49 PM
math2009
$\displaystyle \underset{x\rightarrow 0}{\lim} ~(\cos x)^{\frac{1}{x^2}}= e^{- \frac{1}{2}}=\displaystyle \frac{1}{\sqrt{e}}$

ref attachment
• Jul 9th 2010, 11:56 PM
Also sprach Zarathustra
I forget the ln.... :(

Thank you!
• Jul 10th 2010, 12:21 PM
DeMath
Quote:

Originally Posted by Also sprach Zarathustra
I forget the ln.... :(

Also this limit can be calculated without taking the logarithm and L'Hopital:

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

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

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