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

Statement

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 $\varepsilon$ - $\delta$ 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

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

Then

$\lim_{x \to c} f(x) \;\le\; \lim_{x \to c} g(x) .$

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.

Facts & Assumptions

Given: A set $A \subseteq \mathbb{R}$ , a limit point $c$ of $A$ , functions $f, g : A \to \mathbb{R}$ , reals $L, M$ with $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$ , and a real $\eta > 0$ with $f(x) \le g(x)$ for every $x \in A$ satisfying $0 < |x - c| < \eta$ (The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ , Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ ).

[L1]

The limit condition: for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that every $x \in A$ with $0 < |x - c| < \delta$ satisfies $|f(x) - L| < \varepsilon$ , and likewise for $g$ and $M$ (The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ ).

[L2]

Limit point: for every real $\rho > 0$ there is $x \in A$ with $0 < |x - c| < \rho$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ , The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L3]

Absolute value: for $t > 0$ , $|u| < t$ is equivalent to $-t < u < t$ (Basic properties of the absolute value).

[L4]

Order arithmetic in $\mathbb{R}$ : the order is total, so the negation of $u \le v$ is $v < u$ ; trichotomy, so $u < v$ and $v \le u$ cannot both hold; adding a constant to an inequality and adding two inequalities (Order is preserved by adding a constant and by adding inequalities); $0 < 1$ (The multiplicative identity is positive), so $2 > 0$ , $2^{-1} > 0$ (Inverses of positives are positive, and reciprocation reverses order) and $t/2 > 0$ for $t > 0$ (Sign rules for products and monotonicity of multiplication), with $(t/2) + (t/2) = t$ ; and of finitely many positive reals the smallest is positive (Ordered field).

Proof

technique · contradiction

1.1

Suppose, for contradiction, that $L \le M$ fails; the order being total, this means $M < L$ .

assume-contra

2.1

Then $L - M > 0$ , so $\varepsilon := (L - M)/2 > 0$ , and $L - \varepsilon = (L + M)/2 = M + \varepsilon$ .

step 1.1L4

3.1

By [L1] fix reals $\delta_1, \delta_2 > 0$ such that every $x \in A$ with $0 < |x - c| < \delta_1$ has $|f(x) - L| < \varepsilon$ and every $x \in A$ with $0 < |x - c| < \delta_2$ has $|g(x) - M| < \varepsilon$ ; let $\delta$ be the smallest of $\delta_1$ , $\delta_2$ and $\eta$ , so $\delta > 0$ .

step 2.1L1L4choose

4.1

Since $c$ is a limit point of $A$ , fix $x \in A$ with $0 < |x - c| < \delta$ .

step 3.1L2choose

5.1

That $x$ satisfies $0 < |x - c| < \delta_1$ and $0 < |x - c| < \delta_2$ , so $|f(x) - L| < \varepsilon$ gives $f(x) > L - \varepsilon$ and $|g(x) - M| < \varepsilon$ gives $g(x) < M + \varepsilon$ ; since $L - \varepsilon = M + \varepsilon$ , this yields $g(x) < f(x)$ .

step 3.1step 4.1L3L4

6.1

But that same $x$ satisfies $0 < |x - c| < \eta$ , so the hypothesis gives $f(x) \le g(x)$ , which together with $g(x) < f(x)$ contradicts trichotomy.

step 3.1step 5.1L4

7.1

The assumption that $L \le M$ fails is therefore untenable, and $\lim_{x \to c} f(x) = L \le M = \lim_{x \to c} g(x)$ .

step 6.1L4discharge-contradiction∎

Remarks

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 $\lim_{x \to c} f(x) = L \ne 0$ then $|f| > |L|/2$ on a punctured neighbourhood of $c$ ; in particular if $L > 0$ then $f > L/2 > 0$ 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 $0$ .

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 32 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Lebl, Basic Analysis I, §3.1: Limits of functions (standard reference, not scraped)
Limit of a function (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.1 (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §9.3 (standard reference, not scraped)

If f≤gf \le gf≤g on a punctured neighbourhood of ccc then lim⁡f≤lim⁡g\lim f \le \lim glimf≤limg, non-strictly