First, let #y=(sin(x))^{sin(x)}#. Then #ln(y)=sin(x)ln(sin(x))=(ln(sin(x)))/csc(x)#. Now use L'Hopital's Rule to evaluate the limit of this expression (it is an #infty/infty# indeterminate form).
# lim_{x->0+} (ln(sin(x)))/csc(x)=lim_{x->0+}(1/sin(x) * cos(x))/(-cot(x)*csc(x))#
#=lim_{x->0+}(-tan(x))=0#
Therefore, #ln(lim_{x->0+}y)=lim_{x->0}ln(y)=lim_{x->0+}sin(x)ln(sin(x))=0#. Exponentiation now implies that
#lim_{x->0+}(sin(x))^{sin(x)}=lim_{x->0+}y=e^{0}=1#.
The graph of #(sin(x))^{sin(x)}# confirms this visually.
graph{sin(x)^(sin(x)) [-4.054, 4.065, -2.034, 2.025]}
