The answer is: #2#.
First of all we have to make a clarification: the limit doesn't exist for #xrarr0#, but only for #xrarr0^+# because the second part of the limit, #(1/x)^sinx#, exixts only with positive #x#.
The first part of the limit:
#lim_(xrarr0^+)sinx/x=1# because it is a fundamental limit.
The previous limit also says us that in a neighbourhood of zero #sinx~x#, so the second limit is:
#lim_(xrarr0^+)(1/x)^sinx=lim_(xrarr0^+)(1/x)^x=(+oo)^(0^+)#,
that is an indecision form that we have to transform in another, usinig the property #f(x)=e^lnf(x)#:
#lim_(xrarr0^+)(1/x)^x=lim_(xrarr0^+)x^(-x)=lim_(xrarr0^+)e^(lnx^(-x))=#
#=lim_(xrarr0^+)e^(-xlnx)=lim_(xrarr0^+)e^(-lnx/(1/x))=e^(-(+oo)/(+oo))=#
#=e^(-0^+)=1#.
(The last limit is #0^+# because the infinite of a power of #x# is greater then the infinite of a power of a logarithm).
So the limit is: #1+1=2#.
