Find the value of $\displaystyle \displaystyle\lim_{x\to\infty}({\frac{\pi}{2}-\arctan\:x})^{\frac{1}{ln\:x}}$.
Thanks!
Well, the arctan(x) will go to $\displaystyle \frac{/pi}{2}$ as x goes to infinity, because the range is $\displaystyle -\frac{\pi}{2}, \frac{\pi}{2}$
So the inside goes to $\displaystyle \frac{\pi}{2}-\frac{\pi}{2}=0$
And the power will go to 0, because $\displaystyle ln(\infty)=\infty$ and $\displaystyle \frac{1}{\infty} =0$.
So, yes, it does go to 0^0
OMG you noobs...no it doesn't ok! The answer should be $\displaystyle \frac{1}{e}$. 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.
log 0 and log x are both infinite so we can use L'hopital's rule.
$\displaystyle \frac {d}{dx} (log(\pi/2-arctan x)) = \frac {1/(x^2+1)}{\pi/2-arctan x}$
$\displaystyle \frac {d}{dx} (log x) = 1/x$
so we have $\displaystyle
\frac {\frac {1/(x^2+1)}{\pi/2-arctan x}}{1/x}$
which simplifies to $\displaystyle
\frac {x/(x^2+1)}{\pi/2-arctan x}$
since this is 0/0 we can use l'Hopital's rule again":
$\displaystyle
\frac {d}{dx}(\frac {x}{x^2+1} = \frac {x^2+1-2x^2}{(x^2+1)^2}
=\frac {-x^2+1}{(x^2+1)^2}$
Edit: an angel found my mistake here and I have now fixed it
$\displaystyle \frac {d}{dx} (\pi/2-arctan x) = \frac {-1}{x^2+1}$
so we have
$\displaystyle \frac {\frac {-x^2+1}{(x^2+1)^2}}{\frac{-1}{x^2+1}}$
which simplifies to
$\displaystyle \frac {x^2-1}{x^2+1}$ -> 1
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?