How do you prove that the limit of #x^2 = 0# as x approaches 0 using the epsilon delta proof?

1 Answer
Jun 21, 2016

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#