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.
∎
