To prove that #lim_(x->c)f(x)=L# for some function #f#, we must show that for any #epsilon > 0# there exists a #delta > 0# such that if #|x-c| < delta# then #|f(x)-L| < epsilon|#

**Proof:**

Let #epsilon > 0# be arbitrary, and let #delta = min(1,3epsilon)#. Then, if #|x-1| < delta#, note that #|x-1| < 1# and thus #|2x + 1| > 1# (see below for details). With that, we have, for #|x-1|< delta#:

#|(x+1)/(2x+1)-2/3| = |(3x+3)/(3(2x+1))-(4x+2)/(3(2x+1))|#

#=|(-x+1)/(3(2x+1))|#

#=|(x-1)/(3(2x+1))|#

#=|x-1|/(3|2x+1|)#

#<|x-1|/3#

#<(3epsilon)/3#

#=epsilon#

Therefore, as we have shown that a #delta# exists for any #epsilon# such that #|x-1|< delta|# implies #|(x+1)/(2x+1)-2/3| < epsilon#, it is the case that #lim_(x->1)(x+1)/(2x+1)=2/3" "#∎

To see where we got #|2x+1|< 1#, starting from #|x-1|< 1# we have:

#|x-1| < 1#

#=> -1 < x-1 < 1#

#=> -2 < 2x - 2 < 2#

#=> 1 < 2x+1 < 5#

#=> 1 < |2x + 1| < 5#

#:. |2x+1| > 1#