Evaluate the Limit

• December 5th 2009, 12:59 AM
zorro
Evaluate the Limit
Question : Evaluate

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

Originally Posted by zorro
Question : Evaluate

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

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
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)