limit as x approaches pi/2 of (tan x)^(x-pi/2) I don't even know where to start.
$\displaystyle \left(\tan x\right)^{x-\frac{\pi}{2}}=e^{\left(x-\frac{\pi}{2}\right)\ln\tan x}$ , so using continuity of the exponential formula, apply L'Hospital's rule to:
$\displaystyle \lim_{x\to \pi\slash 2}\left(x-\frac{\pi}{2}\right)\ln\tan x=\lim_{x\to \pi\slash 2}\frac{\ln\tan x}{\frac{1}{x-\pi\slash 2}}$ .
Tonio
well you have tan(pi/2)^0, correct? Well tan is sin/cos, and cos of pi/2 is 0, which would make it undefined. But it's APPROACHING pi/2, not at it. look at it graphically, pi/2 is an asymptote of tanx and to the left of the asymptote f(x) approaches infinity.
So you're really looking at [sin(pi/2)/cos(pi/2)]^0 or [1/inf]^0 or 0^0. This is an indeterminite form of a limit and I don't know if you've learned L'Hopitals rule yet but it needs to be applied here.
We write y=tanx^(x-pi/2) and take the natural log of both sides. lny=(x-pi/2)lntanx we rewrite this to get it to the form 0/0 so we can l'hop it.
lny= lntanx/(x-pi/2)^-1 which is 0/0 as x>pi/2.
now we take the derivative of top and bottom to lhop it.
Well, yes: i think you should know it, but it never minds: you'll now it if you continue reading.
For any $\displaystyle a\,,\,b\in\mathbb{R}\,,\,a>0\,,\,\,a^b=e^{b\ln a}$ . This follows at once from the very definition of logarithm which I urge to read (or hopefully: re-read) and re-understand.
Tonio