What is Limit at a point? — Mathematics

Definition 21.2 High School Mathematics · 第 21 章 — Limits and Continuity

Let $a \in \R$ and let $f$ be defined near $a$ (except possibly at $a$ ). We say that $f$ tends to $\ell \in \R$ at $a$ , written $\lim\limits_{x \to a} f(x) = \ell$ , if for every $\varepsilon > 0$ there exists $\delta > 0$ such that for all $x \in D$ with $\abs{x - a} \leq \delta$ , $\abs{f(x) - \ell} \leq \varepsilon$ .

We say $f$ tends to $+\infty$ at $a$ if for every $M$ there is $\delta > 0$ such that $f(x) \geq M$ whenever $x \in D$ and $\abs{x - a} \leq \delta$ . One-sided limits ( $x \to a^+$ , $x \to a^-$ ) restrict $x$ to $x > a$ or $x < a$ .

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Definition 13.1 University Mathematics — Year 1 · 第 13 章 — Limits and Continuity

Let $x_0 \in \overline{I}$ and $\ell \in \R$ . Then $f(x) \to \ell$ as $x \to x_0$ when

\forall \varepsilon > 0,\ \exists \delta > 0,\ \forall x \in I, \qquad \abs{x - x_0} \leq \delta \implies \abs{f(x) - \ell} \leq \varepsilon .

Limits at $\pm\infty$ and infinite limits are defined by the same pattern ( $\abs{x - x_0} \leq \delta$ becomes $x \geq M$ ; $\abs{f(x) - \ell} \leq \varepsilon$ becomes $f(x) \geq M'$ ). One-sided limits restrict $x$ to $x > x_0$ (written $x \to x_0^+$ ) or $x < x_0$ . The limit is unique when it exists (same proof as Proposition 11.4).

示例

Example 13.2 (A limit through the squeeze)

Compute $\lim_{x \to 0} x\,\bigl\lfloor \frac1x \bigr\rfloor$ . The floor bracketing $\frac1x - 1 < \lfloor \frac1x \rfloor \leq \frac1x$ gives, after multiplying by $x$ (mind the sign!):

1 - x < x\Bigl\lfloor \frac1x \Bigr\rfloor \leq 1 \quad (x > 0), \qquad 1 \leq x\Bigl\lfloor \frac1x \Bigr\rfloor < 1 - x \quad (x < 0),

and both one-sided squeezes close on $1$ : the limit is $1$ . Note what happened: $\lfloor \frac1x\rfloor$ alone has wild jumps near $0$ , but the factor $x$ tames each jump ( $x$ times a unit jump is small), and only the bracketing survives. The closing insight: limits of products of a small factor with a bounded-oscillation factor are squeeze problems, never operation-theorem problems — the operations theorem needs both factors to converge.

Example 13.5 (Why the composition proviso exists)

Let $g(y) = 0$ for $y \neq 0$ and $g(0) = 1$ , and let $f$ be identically $0$ . Then $f(x) \to 0$ as $x \to 0$ , and $g(y) \to 0$ as $y \to 0$ ; yet $g(f(x)) = g(0) = 1$ for every $x$ , so $g \circ f \to 1 \neq 0$ . The inner function sits exactly on the forbidden value $\ell = 0$ forever, and $g$ ’s limit at $0$ ignores what $g$ does at $0$ . The proviso of Corollary 13.4 — either $g(\ell) = m$ (i.e. $g$ continuous at $\ell$ ), or $f \neq \ell$ near $x_0$ — is precisely what rules this out. The closing insight: in practice one composes continuous functions and the proviso is free; it bites only when limits are taken along punctured neighborhoods, which is why the definition of $\lim_{x \to x_0}$ used in this book includes the point when it is in the domain.

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