Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 14

lim_(x->(pi/2)^-)(tan(x))^cos(x)
=lim_(x->(pi/2)^-)e^(cos(x)lntan(x))
applying the limit chain rule,
lim_(x->(pi/2)^-)cos(x)lntan(x)
=lim_(x->(pi/2)^-)(lntan(x)/(1/cos(x)))
Apply L'Hospital rule, Test L'Hospital condition:
=lim_(x->(pi/2)^-)((lntan(x))')/((1/cos(x))')
=lim_(x->(pi/2)^-)((sec^2(x))/tan(x))/(sin(x)/(cos^2(x)))
=lim_(x->(pi/2)^-)(sec^2(x)cos^2(x))/(tan(x)sin(x))
=lim_(x->(pi/2)^-)1/(tan(x)sin(x))
plug in the value to evaluate limit
=1/(tan(pi/2)sin(pi/2))
=1/(oo*1)=1/oo=0
:.lim_(x->(pi/2)^-)e^(cos(x)lntan(x))=lim_(x->(pi/2)^-)e^0=1