We say that the limit as #x->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 #|f(x)-L|< epsilon#.
Using the above definition, the structure of the proof is to take an arbitrary #epsilon > 0#, and then demonstrate the existence of such a #delta#.
Proof:
Let #epsilon > 0# be arbitrary. Let #delta = epsilon#. Then for any #x# such that #0 < |x-2| < delta#, we have
#|(x^2+x-6)/(x-2)-5| = |((x-2)(x+3))/(x-2)-5|#
#=|x+3-5|#
#=|x-2| < delta = epsilon#
Thus, we have shown that for any #epsilon > 0# there exists a #delta > 0# such that #0 < |x-2| < delta# implies #|(x^2+x-6)/(x-2)-5| < epsilon#, meaning #lim_(x->2)(x^2+x-6)/(x-2) = 5#
∎
Note that in the above proof, we have the restriction that #|x-2| > 0#, allowing us to cancel the #x-2# in the denominator without a problem.
