One-sided limit epsilon delta proof

Question

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?

Answer 1

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 $

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

Answer 2

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: $0<x-a<\delta$, 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.

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Here's an example proof for the case you asked about: