If $f\le g$ on a punctured neighbourhood of $c$ then $\lim f\le \lim g$, non-strictly

Let $A\subseteq \mathbb{R}$, let $c$ be a limit point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$), let $f,g:A\to \mathbb{R}$ and suppose both limits at $c$ exist (The $\epsilon$-$\delta$ limit ${\lim }_{x\to c}f(x)=L$ of $f:A\to \mathbb{R}$ at a limit point $c$ of $A$). Suppose further that there is a real $\eta >0$ with

$f(x)\le g(x)\text{for every }x\in A\text{ with }0<\mathrm{\mid }x-c\mathrm{\mid }<\eta \mathrm{.}$

Then

${\lim }_{x\to c}f(x)\le {\lim }_{x\to c}g(x)\mathrm{.}$

The conclusion is non-strict even when the hypothesis is strict. Replacing $\le$ by $<$ on both sides gives a false statement, refuted by FALSE: $f<g$ near $c$ implies $\lim f<\lim g$: strictness is destroyed in the limit, and no hypothesis short of a uniform gap restores it.

Only the values near $c$ matter, by The limit at $c$ depends only on the restriction of $f$ to a punctured neighbourhood of $c$, and passes to any subset of the domain having $c$ as a limit point: the hypothesis is imposed on a punctured neighbourhood of $c$ and on nothing else, and it says nothing about $f(c)$ and $g(c)$, which the definition ignores in any case.

• Both limits are assumed to exist. Nothing here proves existence: the statement compares two numbers that are given. The theorem that produces a limit from an order hypothesis is the squeeze theorem If $f\le g\le h$ near $c$ and $f$ and $h$ have the same limit at $c$, then so does $g$, whose conclusion is exactly the existence of the middle limit.

• The special case $f\equiv 0$ says that a function which is non-negative near $c$ has a non-negative limit there. Its contrapositive is the form used in practice: a negative limit forces negative values near $c$, which is a weak version of If ${\lim }_{x\to c}f(x)=L\mathrm{\ne }0$ then $\mathrm{\mid }f\mathrm{\mid }>\mathrm{\mid }L\mathrm{\mid }\mathrm{/}2$ on a punctured neighbourhood of $c$; in particular if $L>0$ then $f>L\mathrm{/}2>0$ there.

• The sequential analogue is Limits preserve non-strict inequalities, and it too is non-strict for the same reason: the counterexample is a strict inequality between quantities whose difference tends to $0$.