Given a function #f(x)#, we say that the limit as #x# approaches #a# of #f(x)# is #L#, denoted #lim_(x->a)f(x) = L#, if for every #epsilon > 0# there exists a #delta > 0# such that #0 < |x-a| < delta# implies that #|f(x) - L| < epsilon #.
In more intuitive terms, we say that #lim_(x->a)f(x)=L# if we can make #f(x)# arbitrarily "close" to #L# by making #x# close enough to #a#.
Now, to use this in a proof with #f(x) = x^2, a = 2#, and #L = 4#:
Proof: Let #epsilon > 0# be arbitrary. Let #delta = sqrt(epsilon+4)-2# (note that #delta > 0# as #sqrt(epsilon+4) > sqrt(4) = 2#).
Suppose #|x-2| < delta#. Then
#-delta < x-2 < delta#
#=> -delta + 4 < x+2 < delta + 4#
#=> -delta - 4 < x + 2 < delta + 4#
#=> |x+2| < delta + 4#
With that, then if #0 < |x-2| < delta#, we have
#|x^2 - 4| = |x-2| * |x+2|#
# < delta(delta+4)#
# = (sqrt(epsilon+4)-2)(sqrt(epsilon+4)-2+4)#
#=(sqrt(epsilon+4))^2-2^2#
#=epsilon#
We have shown that for any #epsilon > 0# there exists a #delta > 0# such that #0 < |x-2| < delta# implies #|x^2-4| < epsilon#. Thus, by the #epsilon-delta# definition of a limit, #lim_(x->2)x^2 = 4#.
∎
