# One-sided limit epsilon delta proof

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?

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 $$3 $$(0 implies

$$|\dfrac{x+1}{x+2}-5/4| \lt$$

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