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.
