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If $f$ has a finite limit at $c$ then $f$ is bounded on some punctured neighbourhood of $c$

Statement

Let $A \subseteq R$ , let $c$ be a limit point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ), let $f : A \to R$ and suppose the limit of $f$ at $c$ exists, say $\lim_{x \to c} f (x) = L$ (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ). Then there are a real $δ > 0$ and a real $M \geq 0$ with

$∣ f (x) ∣ \leq M for every x \in A with 0 < ∣ x - c ∣ < δ;$

equivalently, the image $f (A \cap N_{δ}^{*} (c))$ is a bounded subset of $R$ (Lower bound, bounded below, bounded set, The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ). One may take $M = ∣ L ∣ + 1$ .

Only local boundedness follows, never boundedness on $A$ . A function with a limit at $c$ may be unbounded on its domain, as FALSE: a function with a limit at $c$ is bounded on its whole domain records.

Facts & Assumptions

Given: A set $A \subseteq R$ , a limit point $c$ of $A$ , a function $f : A \to R$ and a real $L$ with $\lim_{x \to c} f (x) = L$ (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ , Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ).

[L1]

The limit condition: for every real $ε > 0$ there is a real $δ > 0$ such that every $x \in A$ with $0 < ∣ x - c ∣ < δ$ satisfies $∣ f (x) - L ∣ < ε$ (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ).

[L2]

Absolute value: $∣ u ∣ \geq 0$ ; $- ∣ u ∣ \leq u \leq ∣ u ∣$ ; and for $t > 0$ , $∣ u ∣ \leq t$ is equivalent to $- t \leq u \leq t$ (Basic properties of the absolute value).

[L3]

Triangle inequality: $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ (The triangle inequality).

[L4]

Order arithmetic: $0 < 1$ (The multiplicative identity is positive); adding a constant preserves the order and adding inequalities is legitimate (Order is preserved by adding a constant and by adding inequalities); and $u < v$ implies $u \leq v$ . Order is preserved by adding a constant and by adding inequalities states these moves in their STRICT forms only; the non-strict forms used below follow by adjoining the equality case, in which the two sides coincide, the order being total (Ordered field).

[L5]

Bounded set: $S \subseteq R$ is bounded when it has both an upper and a lower bound (Lower bound, bounded below, bounded set); and $N_{δ}^{*} (c) = {y : 0 < ∣ y - c ∣ < δ}$ (The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

Proof

technique · direct

1.1

Apply [L1] with the particular value $ε = 1$ , legitimate since $1 > 0$ : fix a real $δ > 0$ such that every $x \in A$ with $0 < ∣ x - c ∣ < δ$ satisfies $∣ f (x) - L ∣ < 1$ .

L1L4choose

1.2

Put $M : = ∣ L ∣ + 1$ . Then $M \geq 0$ , since $∣ L ∣ \geq 0$ and $1 > 0$ .

L2L4

2.1

For every $x \in A$ with $0 < ∣ x - c ∣ < δ$ we have $∣ f (x) ∣ = ∣ (f (x) - L) + L ∣ \leq ∣ f (x) - L ∣ + ∣ L ∣ < 1 + ∣ L ∣ = M$ , hence $∣ f (x) ∣ \leq M$ .

step 1.1step 1.2L2L3L4

3.1

Therefore $- M \leq f (x) \leq M$ for every such $x$ , so $M$ is an upper bound and $- M$ a lower bound of the image $f (A \cap N_{δ}^{*} (c))$ : that image is a bounded subset of $R$ .

step 2.1L2L5∎

Remarks

The radius $δ$ depends on $f$ and $c$ and on nothing else here, since the proof runs the limit condition at the single value $ε = 1$ . Any other positive $ε$ would do, with $M = ∣ L ∣ + ε$ ; the value $1$ is chosen only because it is available in every ordered field (The multiplicative identity is positive).
Why this is needed. It is the hypothesis that makes the product case of Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero work: to estimate $f (x) g (x) - L M$ one needs a bound on $∣ f (x) ∣$ near $c$ , and the limit of $f$ supplies one only locally. The companion fact for the denominator of a quotient — a lower bound on $∣ g ∣$ near $c$ — is If $\lim_{x \to c} f (x) = L \neq 0$ then $∣ f ∣ > ∣ L ∣ / 2$ on a punctured neighbourhood of $c$ ; in particular if $L > 0$ then $f > L / 2 > 0$ there.
The converse fails. A function bounded on a punctured neighbourhood of $c$ need not have a limit at $c$ : the oscillator $ψ (1 / x)$ on the companion page takes only values in $[0, 1 / 2]$ and has no limit at $0$ .

Depends on

Used by

Dependency tree · two levels

20 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on 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 ff has a finite limit at cc then ff is bounded on some punctured neighbourhood of cc