ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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$(3x^2 - 1)/(x^2 + x) \to 3$ as $x \to +\infty$

Example

Let $A := (0, \infty)$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length) and let

$f : A \to \mathbb{R}, \qquad f(x) := \frac{3x^2 - 1}{x^2 + x}$

(Integer powers $a^m$ ). Then $A$ is not bounded above (Lower bound, bounded below, bounded set), so the limit at $+\infty$ is well posed (Limits at $+\infty$ and $-\infty$ , and infinite limits at a point); it exists, and

$\lim_{x \to +\infty} f(x) \;=\; 3 .$

This is proved by a direct estimate, not by an algebra of limits. Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero is stated at a finite limit point of the domain, and this library proves no algebra of limits at $\pm\infty$ ; the familiar manipulation "divide numerator and denominator by $x^2$ and take limits termwise" is therefore not available here. Instead the whole computation is packed into one inequality, valid for $x \ge 1$ :

$|f(x) - 3| \;=\; \frac{1 + 3x}{x^2 + x} \;\le\; \frac{4}{x} ,$

after which the Archimedean property finishes the argument.

Facts & Assumptions

Given: The set $A = (0,\infty)$ and the function $f(x) = (3x^2 - 1)/(x^2 + x)$ on $A$ .

[L1]

Limits at $+\infty$ : for $A$ not bounded above, $\lim_{x \to +\infty} f(x) = L$ means that for every real $\varepsilon > 0$ there is a real $M$ with $|f(x) - L| < \varepsilon$ for every $x \in A$ with $x > M$ (Limits at $+\infty$ and $-\infty$ , and infinite limits at a point).

[L2]

Archimedean property: for every real $t$ there is a natural $n \ge 1$ with $t < n \cdot 1_{\mathbb{R}}$ ; and for every real $\varepsilon > 0$ there is a natural $n \ge 1$ with $1/n < \varepsilon$ (Every complete ordered field is Archimedean, For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Complete ordered field (least-upper-bound property)). The canonical naturals satisfy $n \cdot 1_{\mathbb{R}} > 0$ and $1 \le n \cdot 1_{\mathbb{R}}$ for $n \ge 1$ , and are increasing in $n$ (Canonical naturals are positive and strictly increasing).

[L3]

Bounded set: $S$ is bounded above when some real is an upper bound of it (Lower bound, bounded below, bounded set); and $(0,\infty) = \{\, x : x > 0 \,\}$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

Order and field arithmetic: products of positives are positive and for $t > 0$ , $u < v$ is equivalent to $ut < vt$ (Sign rules for products and monotonicity of multiplication); $a > 0$ gives $a^{-1} > 0$ and $0 < a < b$ gives $0 < 1/b < 1/a$ , with the non-strict forms following by adjoining equality (Inverses of positives are positive, and reciprocation reverses order); adding inequalities and translation invariance (Order is preserved by adding a constant and by adding inequalities); $0 < 1$ (The multiplicative identity is positive); the field identities (Field); transitivity and totality (Ordered field).

[L5]

Absolute value: $|u| \ge 0$ , $|u| = u$ for $u \ge 0$ , and $|-u| = |u|$ (Basic properties of the absolute value).

[L6]

Powers: $x^2 = x \cdot x$ (Integer powers $a^m$ ).

Verification

technique · direct

1.1

$f$ is defined on all of $A$ : every $x \in A$ has $x > 0$ , hence $x^2 = x \cdot x > 0$ and $x^2 + x > 0$ , so $x^2 + x \ne 0$ and the quotient exists.

L3L4L6

1.2

$A$ is not bounded above: given a real $M$ , [L2] supplies a natural $n \ge 1$ with $M < n \cdot 1_{\mathbb{R}}$ , and $n \cdot 1_{\mathbb{R}} > 0$ puts it in $A$ ; so no real is an upper bound of $A$ , and the limit at $+\infty$ is well posed.

L2L3

2.1

For every $x \in A$ , $f(x) - 3 = \dfrac{(3x^2 - 1) - 3(x^2 + x)}{x^2 + x} = \dfrac{-1 - 3x}{x^2 + x}$ , hence, both $1 + 3x$ and $x^2 + x$ being positive, $|f(x) - 3| = \dfrac{1 + 3x}{x^2 + x}$ .

step 1.1L4L5L6

3.1

For every $x \in A$ with $x \ge 1$ : from $1 \le x$ we get $1 + 3x \le x + 3x = 4x$ , and from $x > 0$ we get $x^2 + x > x^2 > 0$ ; therefore $\dfrac{1 + 3x}{x^2 + x} \le \dfrac{4x}{x^2 + x} \le \dfrac{4x}{x^2} = \dfrac{4}{x}$ , so $|f(x) - 3| \le 4/x$ .

step 2.1L4L6

4.1

Let $\varepsilon > 0$ be an arbitrary real. By [L2] fix a natural $n \ge 1$ with $1/n < \varepsilon$ , and put $M := 4n$ , where $n$ denotes the canonical natural $n \cdot 1_{\mathbb{R}}$ . Since $n \ge 1$ we have $M = 4n \ge 4 > 1$ . For every $x \in A$ with $x > M$ : first $x > 1$ , so step 3.1 applies and $|f(x) - 3| \le 4/x$ ; and $0 < M < x$ gives $0 < 1/x < 1/M$ by [L4], whence $4/x < 4/M = 4/(4n) = 1/n < \varepsilon$ . So $|f(x) - 3| < \varepsilon$ for every $x \in A$ with $x > M$ .

step 3.1L2L4L5

5.1

Since $A$ is not bounded above and for every real $\varepsilon > 0$ such an $M$ has been produced, the limit of $f$ at $+\infty$ exists and equals $3$ .

step 1.2step 4.1L1∎

Remarks

Where the estimate comes from. The exact identity of step 2.1 replaces the informal "the leading terms dominate": it makes $|f(x) - 3|$ a quotient of two explicit positive quantities, and step 3.1 then bounds numerator above and denominator below by the crudest possible expressions, $4x$ and $x^2$ . The constant $4$ is not optimal and does not need to be: the Archimedean property absorbs any constant.
Why the domain is $(0,\infty)$ and not $\mathbb{R}$ . The denominator $x^2 + x$ vanishes at $0$ and at $-1$ , so $f$ is not defined there; restricting to $(0,\infty)$ both makes $f$ a function and makes the denominator positive, which is what lets the absolute values be dropped in step 2.1. Any domain unbounded above and avoiding the two zeros would give the same limit by the same estimate.
The corresponding statement at $-\infty$ would be the limit $3$ on a domain unbounded below and avoiding the two zeros of the denominator, proved from the same identity of step 2.1 with the inequalities on $x$ reversed. It is not asserted here and is not proved here, because nothing on these pages uses it.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 58 results over 16 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

Limit of a function (Wikipedia) (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.5 (standard reference, not scraped)

(3x2−1)/(x2+x)→3(3x^2 - 1)/(x^2 + x) \to 3(3x2−1)/(x2+x)→3 as x→+∞x \to +\inftyx→+∞