# Math Help - Evaluate the Limit

1. ## Evaluate the Limit

Question : Evaluate

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

2. 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.

3. 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 $\frac1{\sqrt e}.$

4. ## Could u please explain me

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

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

6. ## which 2 limits

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...

7. 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.

8. ## there is only on limit

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

,

#Second..... how Krizalid concluded

As we know that and

9. 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.

10. ## Thanks but what about the second one

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

11. 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.

12. ## Is this correct

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??

13. 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 $\lim_{x \to 0} (\cos x)^{\frac{1}{x^2}} = \lim_{x \to 0} e^{\frac{\ln \cos x}{x^2}} = e^{\lim_{x \to 0}{\frac{\ln \cos x}{x^2}}} = ....$

14. ## Thanks u very much

Thank u Mr fantastic and every body else for ur guidance