The limit #lim_(x->0)sin(1/x)# does not exist. You can proove this taking two sequences such as #a_n=1/(2pin)# and #b_n=1/((2n+1)*pi)# and replacing into the value of x.Hence
#sin(1/(1/((2pin))))=sin(2*pi*n)=1 or -1#.Hence the values of the limits both sequences are different such limit does not exist