How do you use the epsilon delta definition to prove that the limit of #x^2-4# as #x->2#?

The #epsilon - delta# (epsilon - delta) definition of a limit is as follows:

The limit of #f(x)# as #x# approaches #a# 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 #|f(x) - L| < epsilon#.

In more intuitive terms, we have #lim_(x->a)f(x) = L# if we can make #f(x)# arbitrarily close to #L# by making #x# close to #a#.

Using that definition, we can show that #lim_(x->2)(x^2-4) = 0#

Proof:

Let #epsilon > 0# be arbitrary, and let #delta = sqrt(4+epsilon)-2#

Note that #delta > 0#, and that if #0 < |x-2| < delta#, we have

#-delta < x-2 < delta#
#=> -delta + 4 < x+2 < delta + 4#
#=> -delta - 4 < x+2 < delta + 4#
#=> |x+2| < delta + 4#

Now, suppose #0 < |x-2| < delta#. Then

#|(x^2-4)-0| = |x^2-4|#

#=|x-2|*|x+2|#

# < delta(delta+4)#

#=delta^2+4delta#

#=(sqrt(4+epsilon)-2)^2+4(sqrt(4+epsilon)-2)#

#=4+epsilon-4sqrt(4+epsilon)+4+4sqrt(4+epsilon)-8#

#=epsilon#

We have shown that #0 < |x-2| < delta# implies #|(x^2-4)-0| < epsilon#, and thus #lim_(x->2)(x^2-4) = 0# by the #epsilon-delta# definition of a limit.
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