Since #ln(x)# is a continuous function for #x>0#, we can take the log of both sides and proceed
# ln(lim_{x to 0+} (1+x)^cot x) = lim_{x to 0+} ln[(1+x)^cot x]#
#qquad = lim_{x to 0+} [cot(x)*ln(1+x)] = lim_{x to 0+}ln(1+x)/tan x#
# qquad = lim_{x to 0+} [ln(1+x)/x*x/sin x* cos x]#
#qquad = lim_{x to 0+}ln(1+x)/x * lim_{x to 0+}x/sin x* lim_{x to 0+}cos x#
#qquad = 1 times 1 times 1 = 1#
Hence
# lim_{x to 0+}(1+x)^cot x = e#
You could also have used l'Hospitals rule to evaluate # lim_{x to 0+}ln(1+x)/tan x#, since it is of the form #0/0#
