We say that the limit of f(x) = L as x approaches a, 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 say that lim_(x->a)f(x) = L if we can make f(x) arbitrarily close (epsilon close) to L by making x close enough (delta close) to a.
Using this definition, a proof with f(x) = x+3, L = 5, and a = 2 may go as follows:
Proof: Let epsilon > 0 be arbitrary. Let delta = epsilon. Then if 0 < |x - 2| < delta, we have
|(x+3) - 5| = |x-2| < delta = epsilon
We have shown that for any epsilon > 0 there exists a delta > 0 such that 0 < |x-2| < delta implies |(x+3)-5| < epsilon, and thus, by definition, lim_(x->2)(x+3) = 5.
∎
Note that our choice of delta may not always be epsilon, but it happens to work in this case. In more complicated examples, it may be helpful to start from |f(x)-L| < epsilon and then manipulate the inequality to find a delta that will make that condition true.
