#lim_(xrarr0^+)sinx^x=?#
#sinx^x=e^(ln(sinx)^x)=e^(xln(sinx))#
#= lim_(xrarr0^+)e^(xln(sinx))# #(1)#
You see that we still have problem and can't calculate the limit at it's current form as #ln0# is not defined.
Let's study this limit seperately
#lim_(xrarr0^+)ln(sinx)=#
Set
#sinx=u#
#x->0^+#
#u->0^+#
#=# #lim_(urarr0^+)lnu=-oo#
graph{lnx [-10, 10, -5, 5]}
#lim_(xrarr0^+)xln(sinx)=lim_(xrarr0^+)ln(sinx)/(1/x)=_(DLH)^(((-oo)/(+oo)))#
#lim_(xrarr0^+)(((sinx)')/sinx)/(-1/x^2)=-lim_(xrarr0^+)(cosx/sinx)/(1/x^2)=-lim_(xrarr0^+)((x^2cosx)/sinx)#
#x->0^+#
#x>0#
#=# #-lim_(xrarr0^+)(cosx)/(sinx/x*1/x)=^((1/(1*(+oo)))# #0#
#lim_(xrarr0^+)cosx=cos0=1#
#lim_(xrarr0^+)sinx/x=1#
#lim_(xrarr0^+)(1/x)=^((1/(0^+))# #+oo#
Now if in #(1)# i set:
#y=xln(sinx)#
#x->0^+#
#y->0#
and #(1)# now becomes:
#lim_(yrarr0)e^y=e^0=1#
As a result,
#lim_(xrarr0^+)sinx^x=1#
