The million $ question in calculus..

find the limit of (2-x)^(tan((Pi*x)/2))
lim x-->1

you can apply L Hospital rule if that helps

Here you go.
First remember that x^a = e^(alnx) (Think about it for a minute, and you'll see it is true if x>0)
So:
lim x-->1 (2-x)^(tan((πx)/2)) is the same as lim x-->1 e^(tan((πx)/2)ln(2-x)) and as
lim x-->1 e^( (sin((πx)/2) * ln(2-x)) /cos((πx)/2)) (Sidenote: tan(x)=sinx/cosx)

So what is the value of lim x-->1 (sin((πx)/2) * ln(2-x)) /cos((πx)/2)?

Lets use l’Hospital’s Rule if you insist (since we have 0/0)
lim x-->1 (sin((πx)/2) * ln(2-x)) /cos((πx)/2) =
lim x-->1 ((πx)/2 * cos((πx)/2) * ln(2-x) + 1/(2-x) *(-1)*sin((πx)/2) / (-(πx)/2 * sin((πx)/2))
Don't forget to apply the Power rule and the Chain rule while finding the derivatives of nominator and denominator.

Substitute x=1 to the above limit and you'll get
(π/2*0*0 + 1*1*-1) / (-π/2 *1) = 1/(π/2) = 2/π

So the answer to your limit is e^(2/π)

DFTBA!

P.S. If you don't trust me check the answer here -
http://www.wolframalpha.com/input/?i=lim+x--%3E1+%282-x%29^%28tan%28%28Pi*x%29%2F2%29%29

P.P.S For other "mathematicians" - tan(π/2) is complex infinity the value of tan(π/2) approaching from the right is -∞ and from the left is ∞
Also 1^∞ is not 1 it is indeterminate form! Think before you answer, don't mislead people!

As x approaches 1 the power approaches the tan(pi/2). At pi/2 on the unit circle, the coordinates are (0,1). Since tangent deals with y/x and x=0 at pi/2, the exponent becomes infinity.

However, the base approaches 1. 1 raised to anything,even infinity is 1, so

Therefore, the limit is 1.

(2 - x)^(tan(Pi*x/2)
(2 - 1)^(tan(Pi*1/2)
(1)^(tan(0.5Pi) radian (You have to calculate this in radian, but since I don't have a graphic calculator, I can switch it to degrees by doing time 180 and divided by Pi.)
(1)^(tan(90) degrees
1^-2
1/2
0.5

It is actually exponent -1.995200412208242, but I rounded it to -2.