The #epsilon-delta# definition of a limit states that #lim_(x->a)f(x)=L# if for every #epsilon > 0# there exists #delta > 0# such that #0<|x-a| < delta# implies #|f(x)-L| < epsilon#.
Using this for a proof, then, we start by taking an arbitrary #epsilon > 0#, and then showing that such a #delta# exists.
Proof:
Let #epsilon > 0# be arbitrary, and let #delta = min{1/2, epsilon}#. Note that because #delta <= 1/2# we have #delta^2 < delta <= epsilon#. Additionally, note that #|sin(1/x)|<=1# for all #x in RR#, #x!=0#.
Now, suppose #0 < |x-0| = |x| < delta#. Then
#|x^2sin(1/x)-0| = |x|^2*|sin(1/x)|#
# <= |x|^2 * 1#
# = |x|^2#
# < delta^2#
# < epsilon#
This shows that for an arbitrary #epsilon > 0#, there exists a #delta > 0# such that #0 < |x - 0| < delta# implies #|x^2sin(1/x) - 0| < epsilon#. Thus, by definition, #lim_(x->0)x^2sin(1/x) = 0#
∎
