"g"(x)=(tan(x))^x
ln("g"(x))=xln(tan(x))
x\to \pi/2, tan(x) \to +\infty, and ln(u) \to +\infty as u \to +\infty.
Then as x\to \pi/2,
ln("g"(x)) \to \pi/2 +\infty,
ln("g"(x)) +\infty,
"g"(x) \to +\infty.
"g"(x) diverges as x goes to \pi/2.
Initially, I misread your question as x\to 0. I wrote out a full solution so I thought I may as well leave it here. It's also slightly more interesting than the question actually asked.
"g"(x)=(tan(x))^x
ln("g"(x))=xln(tan(x))
The limit as x\to0 of "g"(x) can now be comptued by writing x=1/(1/x) and using L'Hopital's rule.
Doing so,
lim_{x\to\0} ln(tan(x))/(x^-1) = lim_{x\to\0} (sec^2(x)/tan(x))/(-x^-2),
lim_{x\to\0} ln(tan(x))/(x^-1) = lim_{x\to\0} -(x^2)/(cos^2(x)*sin(x)/(cosx),
lim_{x\to\0} ln(tan(x))/(x^-1) = lim_{x\to\0} -(x^2)/(sin(x)cos(x),
lim_{x\to\0} ln(tan(x))/(x^-1) = lim_{x\to\0} -(2x^2)/(sin(2x)).
The right hand side is still in an indeterminate form so we can apply L'Hopital again.
lim_{x\to\0} -(2x^2)/(sin(2x)) = lim_{x\to0}(-4x)/(2cos(2x)).
The right hand side is computed as 0.
Then,
lim_{x\to0}ln("g"(x))=0.
Applying an inverse logarithm,
lim_{x\to0}"g"(x)=1.
