Im having this problem with a proof for
$\lim_{x \to 3^+} \frac{x+1}{x+2} = \frac{4}{5}$
It should be relatively simple, but im wondering what's the difference between this and the "regular" proof for limit of a function?
This should be the definition for the "regular" epsilon-delta proof:
$\forall \epsilon > 0$ $\exists \delta > 0$ such that, when $ 0 < |x – a| < \delta $, then $|f(x) – L | < \epsilon $
And what i've found is that this should be for the one-sided limit:
$\forall \epsilon > 0$ $\exists \delta > 0$ such that, when $ a < x < a + \delta $, then $|f(x) – L | < \epsilon $
So my question is, how does one build up the proof for these one-sided limits?
Best Answer
Let $x \gt 3$:
$|\dfrac{x+1}{x+2}-4/5| =$
$|\dfrac{5x+5-4x-8}{5(x+2)}|= \dfrac{x-3}{5(x+2)} \lt$
$\dfrac{x-3}{5 \cdot 3}=\dfrac{x-3}{15}.$
Let $\epsilon>0$ be given.
Choose $\delta =15 \epsilon.$
Then $3 <x<3+\delta$ $(0<x-3<\delta)$ implies
$|\dfrac{x+1}{x+2}-5/4| \lt $
$\dfrac{x-3}{15} <\delta/(15)=\epsilon$.