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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?

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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 $

$\dfrac{x-3}{15} <\delta/(15)=\epsilon$.

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