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.
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