let #y=lim_(xrarr0^+)((1/tan(x))^(1/ln(x)))#
take ln() of both sides: #lny=ln[lim_(xrarr0^+)((1/tan(x))^(1/ln(x)))]#
move ln() into the limit: #lny=lim_(xrarr0^+)(ln[(1/tan(x))^(1/ln(x))])#
logarithm properties: #lny=lim_(xrarr0^+)((1/ln(x))ln(1/tan(x)))#
#lny=lim_(xrarr0^+)(ln(1/tan(x))/ln(x))#
if you plug in #x=0^+# directly, you get: #ln(1/tan(0^+))/ln(0^+)#
#=ln(1/0^+)/ln(0^+)#
#=ln(oo)/-oo#
#=oo/-oo#
indeterminate, so use l'hopital's rule:
#lny=lim_(xrarr0^+)((d/dx[ln(1/tan(x))])/(d/dx[ln(x)]))#
#lny=lim_(xrarr0^+)((tan(x)*d/dx(1/tan(x)))/(1/x))#
#lny=lim_(xrarr0^+)(xtan(x)*d/dx(cot(x)))#
#lny=lim_(xrarr0^+)(xtan(x)*(-csc^2(x)))#
#lny=lim_(xrarr0^+)((-xsin(x)/cos(x))/sin^2(x))#
#lny=lim_(xrarr0^+)(-x/(cos(x)sin(x)))#
again, plugging in #x=0^+# results in 0/0 (indeterminate), so use l'hopitals rule again:
#lny=lim_(xrarr0^+)((d/dx[-x])/(d/dx[(cos(x)sin(x))]))#
#lny=lim_(xrarr0^+)(-1/(-sin^2(x)+cos^2(x)))#
now plug in #x=0^+#:
#lny=lim_(xrarr0^+)(-1/(-sin^2(0^+)+cos^2(0^+)))#
#lny=lim_(xrarr0^+)(-1/1)#
#lny=-1#
#y=e^(-1)#
#y=1/e#
you've solved for y, which equals the limit you want to find.
check with this graph: graph{(1/tan(x))^(1/ln(x)) [-1.695, 2.15, -0.3, 1.742]}
as x approaches 0 from the right, the limit is about 0.37, or #1/e#
