Question : Evaluate

$\displaystyle

\lim_{x \to 0} \ (Cos x)^{\frac{1}{x^2}}

$

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- Dec 5th 2009, 12:59 AMzorroEvaluate the Limit
Question : Evaluate

$\displaystyle

\lim_{x \to 0} \ (Cos x)^{\frac{1}{x^2}}

$ - Dec 5th 2009, 01:45 AMmr fantastic
Since $\displaystyle (\cos x)^{\frac{1}{x^2}} = e^{\frac{\ln \cos x}{x^2}}$, consider $\displaystyle \lim_{x \to 0} \frac{\ln \cos x}{x^2}$ (I suggest using l'Hopital's Rule).

If you need more help, please show all your working and state specifically where you're stuck. - Dec 5th 2009, 02:53 AMKrizalid
as $\displaystyle x\to0$ we know that $\displaystyle \frac{\ln(1-x)}x\to-1$ and $\displaystyle \frac{1-\cos x}{x^2}\to\frac12.$

now following up on mr fantastic work, we have that for $\displaystyle x\ne0$ is $\displaystyle \frac{\ln (\cos x)}{x^{2}}=\frac{\ln \big(1-(1-\cos x)\big)}{1-\cos x}\cdot \frac{1-\cos x}{x^{2}},$ and as $\displaystyle x\to0$ the whole limit is $\displaystyle \frac1{\sqrt e}.$ - Dec 6th 2009, 04:00 PMzorroCould u please explain me
- Dec 6th 2009, 04:59 PMDefunkt
Hint: Use l'hopital's on both limits.

- Dec 6th 2009, 06:18 PMzorrowhich 2 limits
- Dec 6th 2009, 06:32 PMmr fantastic
- Dec 6th 2009, 07:11 PMzorrothere is only on limit

Sorry for annoying u so much but i cant understand

#First.... how you got

http://www.mathhelpforum.com/math-he...7a10d6c0-1.gif,

#Second..... how Krizalid concluded

As http://www.mathhelpforum.com/math-he...20901745-1.gif we know that http://www.mathhelpforum.com/math-he...148f4159-1.gif and http://www.mathhelpforum.com/math-he...7346b892-1.gif - Dec 6th 2009, 07:27 PMKrizalid
mmm, that's serious.

first one is a known property, it's just $\displaystyle a=e^{\ln a}.$

and those limits, pretty basic, you can even prove them (if that pleases you) by applying L'Hôpital's rule. - Dec 6th 2009, 07:55 PMzorroThanks but what about the second one
- Dec 7th 2009, 12:38 AMmr fantastic
- Dec 7th 2009, 04:18 PMzorroIs this correct

$\displaystyle \lim_{x \to 0} \frac{ln \ cosx}{x^2}$ = $\displaystyle \lim_{x \to 0} \frac{f(x)}{g(x)}$

Using L'Hospitals Rule weget

$\displaystyle \lim_{x \to 0} \frac{f(x)}{g(x)}$ = $\displaystyle \lim_{x \to 0} \frac{f'(x)}{g'(x)}$

$\displaystyle \lim_{x \to 0} \frac{f'(x)}{g'(x)}$ = $\displaystyle \lim_{x \to 0} \frac{- tanx}{2x}$

Dividing nu & de by x weget

= $\displaystyle \lim_{x \to 0} \frac{- \frac{tanx}{x}}{2}$

= $\displaystyle - \frac{ 1}{2}$

Is this correct or have i done something wrong.....And after this what should i do?? - Dec 8th 2009, 12:51 AMmr fantastic
- Dec 8th 2009, 02:36 AMzorroThanks u very much
Thank u Mr fantastic and every body else for ur guidance(Bow)