1. ## crazy limit

Find the value of $\displaystyle\lim_{x\to\infty}({\frac{\pi}{2}-\arctan\:x})^{\frac{1}{ln\:x}}$.

Thanks!

2. Originally Posted by polymerase
Find the value of $\displaystyle\lim_{x\to\infty}({\frac{\pi}{2}-arctan\:x})^{\frac{1}{ln\:x}}$.

Thanks!
It goes to $0^0$, so it is either 0 or 1.

3. Originally Posted by colby2152
It goes to $0^0$, so it is either 0 or 1.
NO IT DOESN'T

4. Originally Posted by polymerase
NO IT DOESN'T
Well, the arctan(x) will go to $\frac{/pi}{2}$ as x goes to infinity, because the range is $-\frac{\pi}{2}, \frac{\pi}{2}$

So the inside goes to $\frac{\pi}{2}-\frac{\pi}{2}=0$

And the power will go to 0, because $ln(\infty)=\infty$ and $\frac{1}{\infty} =0$.

So, yes, it does go to 0^0

5. Originally Posted by angel.white
Well, the arctan(x) will go to $\frac{/pi}{2}$ as x goes to infinity, because the range is $-\frac{\pi}{2}, \frac{\pi}{2}$

So the inside goes to $\frac{\pi}{2}-\frac{\pi}{2}=0$

And the power will go to 0, because $ln(\infty)=\infty$ and $\frac{1}{\infty} =0$.

So, yes, it does go to 0^0
OMG you noobs...no it doesn't ok! The answer should be $\frac{1}{e}$. trust me this is challenge question my prof asked...and if you knew him you would know his challenge questions are never easy. I may know the answer but i don't know how to get it...and either do you guys.

6. Originally Posted by polymerase
OMG you noobs...no it doesn't ok! The answer should be $\frac{1}{e}$. trust me this is challenge question my prof asked...and if you knew him you would know his challenge questions are never easy. I may know the answer but i don't know how to get it...and either do you guys.
Prove it

7. Originally Posted by angel.white
Prove it
How about you graph it and see....

8. On a graph, yes, I get 1/e. The equation does go to 0^0, but that form isn't differentiable, so it is wrong to say it must be zero or 1.

9. Originally Posted by angel.white
On a graph, yes, I get 1/e. The equation does go to 0^0, but that form isn't differentiable, so it is wrong to say it must be zero or 1.
you learn something everyday don't ya....

see if you find or talk to someone who can figure out the answer....i really would like to know.

10. Originally Posted by polymerase
Find the value of $\displaystyle\lim_{x\to\infty}({\frac{\pi}{2}-arctan\:x})^{\frac{1}{ln\:x}}$.

Thanks!
$\left( \frac{\pi}{2} - \tan^{-1} x \right) ^{ \frac{1}{\log x} } = \exp \left( \frac{\log (\pi/2 - \tan^{-1} x)}{\log x} \right)$
But,
$\frac{\log (\pi/2 - \tan^{-1} x)}{\log x} \to - 1$.
Thus the limit is,
$e^{-1}$.

11. $\frac{\log (\pi/2 - \tan^{-1} x)}{\log x} \to - 1$

TPH: can you show this? I messed with it a while, I kept getting zero (maybe I made an early mistake that kept interfering with my answer)

12. log 0 and log x are both infinite so we can use L'hopital's rule.

$\frac {d}{dx} (log(\pi/2-arctan x)) = \frac {1/(x^2+1)}{\pi/2-arctan x}$

$\frac {d}{dx} (log x) = 1/x$

so we have $
\frac {\frac {1/(x^2+1)}{\pi/2-arctan x}}{1/x}$

which simplifies to $
\frac {x/(x^2+1)}{\pi/2-arctan x}$

since this is 0/0 we can use l'Hopital's rule again":
$
\frac {d}{dx}(\frac {x}{x^2+1} = \frac {x^2+1-2x^2}{(x^2+1)^2}
=\frac {-x^2+1}{(x^2+1)^2}$

Edit: an angel found my mistake here and I have now fixed it
$\frac {d}{dx} (\pi/2-arctan x) = \frac {-1}{x^2+1}$

so we have
$\frac {\frac {-x^2+1}{(x^2+1)^2}}{\frac{-1}{x^2+1}}$

which simplifies to

$\frac {x^2-1}{x^2+1}$ -> 1

13. Thankyou badgerigar, when I followed along, though, I got

$\frac {d}{dx} (\pi/2-arctan x) = -\frac {1}{x^2+1}$

as opposed to positive 1/(x^2+1), so my answer was positive one.

14. Thanks for pointing that out angel-white. I guess there must be some other mistake in there also to make the answer come out right.

I'm new here, so could someone tell me for future reference: If somebody finds a mistake you made is it good or bad to edit your post and fix it?

15. Hmm, I don't know. For me, if the mistake invalidates the entire post, I just put a disclaimer at the top, but leave it there so that the responses to the post make sense. If the mistake is something simple I can fix, then I usually do.

Page 1 of 2 12 Last