Given:
#f(x) = (sinx)^x#
Consider:
#g(x) = ln(f(x)) = ln ((sinx)^x) = x ln sinx#
If we consider the limit:
#lim_(x->0) x ln sinx#
it is in the indeterminate form #0*oo# but we can reduce it to the form #oo/oo# by writing it as:
#g(x) = ln sinx /(1/x)#
and then apply l'Hospital's rule:
#lim_(x->0) g(x) = lim_(x->0) (d/dx( ln sinx))/(d/dx (1/x)) = lim_(x->0) -x^2 cosx/sinx #
As we know that: #lim_(x->0) sinx/x = 1#
we have:
#lim_(x->0) g(x) = lim_(x->0) (-x)(x/sinx) cosx = 0 xx 1 xx 1 = 0 #
Now, as:
#f(x) = e^g(x)#
and as #e^x# is continuous in all #RR# we have:
#lim_(x->0) f(x) = lim_(x->0) e^g(x) = e^(lim_(x->0) g(x)) = e^0 = 1#
