Limit as x approaches 1 of (2 - x)^tan(pix/2)

I know that the limit is undefined at 1 because tan(pi/2) is undefined. It appears to get closer to 1, but I wondered, mathematically, how to evaluate the raised trig function

Well the limit takes the indeterminate form 0^oo. What you can do is use manipulations to bring that exponent down in a legal way. And yes, the log function "ln" is of great help! Supposing that the limit exists and it is L. take ln L. Use the fact that ln is continuous as to switch ln and the limit symbol. C'mon you can do it! Use the following property:

ln(a^r)=r ln a.

Then apply L'Hopital's rule if in case the problem requires.

Since this limit takes the form 1^infinity, we need to let:
L = lim (x-->1) (2 - x)^[tan(πx/2)].

Then, by taking the natural logarithm of both sides:
ln(L) = lim (x-->1) {(2 - x)^[tan(πx/2)]}
= lim (x-->1) tan(πx/2)*ln(2 - x).

This limit now takes the form infinity*0. Re-writing tan(πx/2) as:
1/[1/tan(πx/2)] = 1/cot(πx/2),

gives:
ln(L) = lim (x-->1) tan(πx/2)*ln(2 - x)
= lim (x-->1) ln(2 - x)/cot(πx/2).

This limit now takes the form 0/0, so applying L'Hopital's Rule gives:
ln(L) = lim (x-->1) ln(2 - x)/cot(πx/2)
= lim (x-->1) [-1/(2 - x)]/[(-π/2)csc^2(πx/2)]
= 2/π * lim (x-->1) sin^2(πx/2)/(2 - x)
= (2/π)[1/(2 - 1)], by evaluating at x = 1
= 2/π.

Therefore, ln(L) = 2/π and:
L = lim (x-->1) (2 - x)^[tan(πx/2)] = e^(2/π).
(Verified: http://www.wolframalpha.com/input/?i=lim… )

I hope this helps!