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
