Evaluate the Limit

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December 5th 2009, 12:59 AM
zorro

Evaluate the Limit

Question : Evaluate

$<br /> \lim_{x \to 0} \ (Cos x)^{\frac{1}{x^2}}<br />$
December 5th 2009, 01:45 AM
mr fantastic

Quote:

Originally Posted by zorro

Question : Evaluate

$<br /> \lim_{x \to 0} \ (Cos x)^{\frac{1}{x^2}}<br />$

Since $(\cos x)^{\frac{1}{x^2}} = e^{\frac{\ln \cos x}{x^2}}$ , consider $\lim_{x \to 0} \frac{\ln \cos x}{x^2}$ (I suggest using l'Hopital's Rule).

If you need more help, please show all your working and state specifically where you're stuck.
December 5th 2009, 02:53 AM
Krizalid

as $x\to0$ we know that $\frac{\ln(1-x)}x\to-1$ and $\frac{1-\cos x}{x^2}\to\frac12.$

now following up on mr fantastic work, we have that for $x\ne0$ is $\frac{\ln (\cos x)}{x^{2}}=\frac{\ln \big(1-(1-\cos x)\big)}{1-\cos x}\cdot \frac{1-\cos x}{x^{2}},$ and as $x\to0$ the whole limit is $\frac1{\sqrt e}.$
December 6th 2009, 04:00 PM
zorro

Could u please explain me

Quote:

Originally Posted by Krizalid

as $x\to0$ we know that $\frac{\ln(1-x)}x\to-1$ and $\frac{1-\cos x}{x^2}\to\frac12.$

Could u please explain me how did u jump to the conclusion that $\frac{\ln(1-x)}x\to-1$ and $\frac{1-\cos x}{x^2}\to\frac12.$

I am a novice so if u could provide me with the link where this can be derived i would appreciate it a lot
December 6th 2009, 04:59 PM
Defunkt

Hint: Use l'hopital's on both limits.
December 6th 2009, 06:18 PM
zorro

which 2 limits

Quote:

Originally Posted by Defunkt

Hint: Use l'hopital's on both limits.

could u tell me which two limits ....i can see only 1 limit...
December 6th 2009, 06:32 PM
mr fantastic

Quote:

Originally Posted by zorro

could u tell me which two limits ....i can see only 1 limit...

For goodness sake! Go back and read post #4, your post. The one in which you asked about two limits. That is the post that Defunkt replied to.
December 6th 2009, 07:11 PM
zorro

there is only on limit

Quote:

Originally Posted by mr fantastic

For goodness sake! Go back and read post #4, your post. The one in which you asked about two limits. That is the post that Defunkt replied to.

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
December 6th 2009, 07:27 PM
Krizalid

mmm, that's serious.

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

and those limits, pretty basic, you can even prove them (if that pleases you) by applying L'Hôpital's rule.
December 6th 2009, 07:55 PM
zorro

Thanks but what about the second one

Quote:

Originally Posted by Krizalid

mmm, that's serious.

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

and those limits, pretty basic, you can even prove them (if that pleases you) by applying L'Hôpital's rule.

Thanks for ur reply but what about the second one , could u please tel me how u derived that
December 7th 2009, 12:38 AM
mr fantastic

Quote:

Originally Posted by zorro

Thanks for ur reply but what about the second one , could u please tel me how u derived that

There are several options for finding $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$ but the simplest option is the best one for you to use: Apply l'Hopital's Rule twice.
December 7th 2009, 04:18 PM
zorro

Is this correct

Quote:

Originally Posted by mr fantastic

Since $(\cos x)^{\frac{1}{x^2}} = e^{\frac{\ln \cos x}{x^2}}$ , consider $\lim_{x \to 0} \frac{\ln \cos x}{x^2}$ (I suggest using l'Hopital's Rule).

If you need more help, please show all your working and state specifically where you're stuck.

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

Using L'Hospitals Rule weget

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

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

Dividing nu & de by x weget

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

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

Is this correct or have i done something wrong.....And after this what should i do??
December 8th 2009, 12:51 AM
mr fantastic

Quote:

Originally Posted by zorro

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

Using L'Hospitals Rule weget

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

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

Dividing nu & de by x weget

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

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

Is this correct or have i done something wrong.....And after this what should i do??

Correct.

So $\lim_{x \to 0} (\cos x)^{\frac{1}{x^2}} = \lim_{x \to 0} e^{\frac{\ln \cos x}{x^2}} = e^{\lim_{x \to 0}{\frac{\ln \cos x}{x^2}}} = ....$
December 8th 2009, 02:36 AM
zorro

Thanks u very much

Thank u Mr fantastic and every body else for ur guidance(Bow)