# 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

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

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

##### TL;DR

Only the assumption-you-make/condition-you-have at the beginning changes, namely instead of: $$\vert{x-a}\vert<\delta\Rightarrow\vert{f(x)-L\vert}<\epsilon$$ you would have: $$0<{x-a}<\delta\Rightarrow\vert{f(x)-L\vert}<\epsilon$$ and for a limit as $$x$$ approaches $$a$$ from the left, you, analogously, would have: $$-\delta<{x-a}<0\Rightarrow\vert{f(x)-L\vert}<\epsilon$$

i.e you do the exact same as you would for a two-sided limit, except that instead of assuming $$\vert{x-a}\vert<\delta \iff -\delta < x-a < \delta \iff \delta\in (a-\delta,a+\delta)\backslash \{a\}$$, you would just care about the range of inputs to the right of $$a$$, thus only assuming: $$0, and then using that somehow to find a value for $$\delta$$, and to find bounds for the factors and terms in the function if required, etc. 