We must show that #lim_{x->0}f(x)=f(0)=0#.

Method 1 (use the #epsilon/delta# definition of a limit):

Let #epsilon>0# be given (this represents an arbitrarily small distance that we'd like the function outputs to be to the limit 0 if #x# is sufficiently close to 0).

Choose #delta=sqrt(epsilon)>0# (this represents the measure of "sufficiently close" and is chosen this way to make the algebra below work out nicely).

Suppose #|x-0| < delta# so that #-sqrt(epsilon) < x < sqrt(epsilon)#. Then #0 leq x^{2} < (sqrt(epsilon))^{2}=epsilon# and #|x^{2}sin(1/x)| < epsilon# (when #x!=0#), since #-1 <= sin(1/x) <= 1# for all #x!=0# (also note that #|f(0)|=0<epsilon#).

But this means that #|f(x)-0|< epsilon# for all #x# such that #|x-0| < delta# and we have shown that #lim_{x->0}f(x)=f(0)=0#, making #f# continuous at #x=0#.

Method 2 (use the Squeeze Theorem ):

The facts that #-1 <= sin(1/x) <=1# for all #x!=0# and #f(0)=0# imply that #-x^{2} <= f(x) <= x^{2}# for all #x#. Since #lim_{x->0}x^{2}=0# and #lim_{x->0}(-x^{2})=0#, it follows that #lim_{x->0}f(x)=0=f(0)# by the Squeeze Theorem.