The function, as given, is not continuous at #0# as #0sin(1/0)# is not defined. However, we may make a slight modification to make the function continuous, defining #f(x)# as

#f(x) = {(xsin(1/x)" if "x!=0),(0" if "x=0):}#

We will proceed using this modified function.

Using the #epsilon-delta# definition of a limit, we must show that for any #epsilon > 0# there exists a #delta > 0# such that if #0 < |x - 0| < delta# then #|f(x) - f(0)| < epsilon#

To do so, we first let #epsilon > 0# be arbitrary. Next, let #delta = epsilon/2#. Now, suppose #0 < |x-0| = |x| < delta#. Note that as #|x| > 0# we have #f(x) = xsin(1/x)#. Proceeding,

#|f(x) - f(0)| = |xsin(1/x) - 0|#

#=|xsin(1/x)|#

#=|x|*|sin(1/x)|#

#<=|x|" "# (as #|sin(1/x)|<=1# for all #x !=0#)

#< delta#

#=epsilon/2#

#< epsilon#

So, as for an arbitrary #epsilon > 0# there exists a #delta > 0# such that if #0 < |x-0| < delta# then #|f(x)-f(0)| < epsilon#, we have shown that #f(x)# is continuous at #x=0#.