A function #f# is said to have a limit #L# as #x# approaches #c#, denoted #lim_(x->c)f(x) = L#, if for every #epsilon>0#, there exists a #delta > 0# such that #|x-c| < delta# implies #|f(x)-L| < epsilon#.
Then, to prove that #lim_(x->0)x^2=0#, we must show that for any #epsilon > 0# there exists #delta > 0# such that #|x-0| < delta# implies #|x^2-0| < epsilon#.
Proof:
Let #epsilon > 0# be arbitrary, and let #delta = min{epsilon,1}#.
Suppose #|x-0|=|x| < delta#. Note that as #delta <= 1#, we have #|x| <1#, meaning #|x^2|=|x|^2<|x|*1=|x|#. Furthermore, as #delta <= epsilon#, we have #|x| < epsilon#. With those facts:
#|x^2-0| = |x^2| = |x|^2 < |x| < epsilon#
Thus, for an arbitrary #epsilon > 0#, we have found a #delta > 0# such that #|x-0| < delta# implies #|x^2-0| < epsilon#, meaning #lim_(x->0)x^2=0#
∎
