#"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#.
