Determining Limits Using the Squeeze Theorem · 利用夹逼定理求极限
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| Squeeze Theorem/skwiːz ˈθɪərəm/ | 夹逼定理 | jiā bī dìng lǐ |
| inequality/ɪniːˈkwɒlɪti/ | 不等式 | bù děng shì |
Trap a wild function between two tame ones
- Some limits resist substitution and algebra — like $\displaystyle\lim_{x\to0}x^2\sin\tfrac1x$, which wiggles infinitely fast.
- The escape: trap the messy function between two friendlier ones.
- If the upper and lower "walls" both head to the same height, the trapped function has nowhere else to go.
- This is the Squeeze Theorem 夹逼定理 (also called the Sandwich Theorem).
把一个"狂野"函数夹在两个"温顺"函数之间
- 有些极限既抵抗代入又抵抗代数——比如 $\displaystyle\lim_{x\to0}x^2\sin\tfrac1x$,它无限快地摆动。
- 出路:把这个杂乱的函数夹在两个更友好的函数之间。
- 如果上下两堵"墙"都奔向同一高度,被夹的函数就无处可逃。
- 这就是夹逼定理(也叫三明治定理)。
Trapping a function between two others with equal limits uses the ____ Theorem. · 将一个函数限制在两个极限相等的其他函数之间使用的是 ____ 定理。
Also called the Sandwich Theorem. · 也称为三明治定理。
What the theorem says
- Suppose near $c$ (except maybe at $c$ itself) we have $g(x)\le f(x)\le h(x)$.
- If $\displaystyle\lim_{x\to c}g(x)=\lim_{x\to c}h(x)=L$, then $\displaystyle\lim_{x\to c}f(x)=L$ too.
- The two bounds squeeze $f$ onto the single value $L$.
- You need two things: a valid inequality 不等式 near $c$, and equal limits for the bounds.
定理说了什么
- 设在 $c$ 附近(也许 $c$ 本身除外)有 $g(x)\le f(x)\le h(x)$。
- 若 $\displaystyle\lim_{x\to c}g(x)=\lim_{x\to c}h(x)=L$,则 $\displaystyle\lim_{x\to c}f(x)=L$ 也成立。
- 两个界把 $f$ 挤到那唯一的值 $L$ 上。
- 你需要两样东西:$c$ 附近成立的不等式,以及两个界相等的极限。
To apply the Squeeze Theorem to $f$ near $c$, you need all · 所有 of... · 要对 $f$ 在 $c$ 附近应用夹逼定理，你需要所有的……
You need two-sided bounds with equal limits, valid near $c$. $f$ need not be a polynomial — that is the whole point. · 你需要两侧界限具有相等的极限，且在 $c$ 附近有效。$f$ 不必是多项式 — 这正是该定理的核心所在。
The classic wiggle
- $\sin\tfrac1x$ swings between $-1$ and $1$ forever as $x\to0$ — no limit on its own.
- But multiply by $x^2$: since $-1\le\sin\tfrac1x\le1$, we get $-x^2\le x^2\sin\tfrac1x\le x^2$.
- Both walls $\pm x^2 \to 0$ as $x\to0$.
- Squeezed between them, $\displaystyle\lim_{x\to0}x^2\sin\tfrac1x=0$.
经典的摆动
- 当 $x\to0$,$\sin\tfrac1x$ 永远在 $-1$ 与 $1$ 之间摆动——它自己没有极限。
- 但乘上 $x^2$:因为 $-1\le\sin\tfrac1x\le1$,得到 $-x^2\le x^2\sin\tfrac1x\le x^2$。
- 当 $x\to0$,两堵墙 $\pm x^2 \to 0$。
- 被夹在中间,$\displaystyle\lim_{x\to0}x^2\sin\tfrac1x=0$。
The squeezing wall $y=x^2$ · 夹逼墙 $y=x^2$
y = ax²
The parabola $y=x^2$ and its mirror $-x^2$ both pinch to $0$ at the origin — anything trapped between them is forced to $0$. · 抛物线 $y=x^2$ 及其镜像 $-x^2$ 在原点处共同挤压至 $0$ — 被夹在两者之间的任何函数都被迫趋向于 $0$。
Using $-x^2\le x^2\sin\tfrac1x\le x^2$, find $\displaystyle\lim_{x\to0}x^2\sin\tfrac1x$. · 利用 $-x^2\le x^2\sin\tfrac1x\le x^2$ 求 $\displaystyle\lim_{x\to0}x^2\sin\tfrac1x$。
Both walls $\pm x^2\to 0$, so the trapped function $\to 0$. · 两侧墙壁均趋向于 $\pm x^2\to 0$，因此被夹住的函数趋向于 $\to 0$。
Checking the setup honestly
- The inequality only has to hold near $c$, not everywhere — a small interval around $c$ is enough.
- The bounds' limits must be equal; if the walls head to different heights, the theorem says nothing.
- Squeeze is the standard route to the famous $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$, from the geometric bounds $\cos x \le \frac{\sin x}{x}\le 1$.
- Always name your two bounding functions explicitly — that is the heart of the argument.
诚实地检查前提
- 不等式只需在 $c$ 附近成立,而非处处成立——$c$ 周围一个小区间就够了。
- 两个界的极限必须相等;若两墙奔向不同高度,定理什么都不说。
- 夹逼是求那个著名极限 $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$ 的标准途径,用几何界 $\cos x \le \frac{\sin x}{x}\le 1$。
- 永远明确写出你的两个界函数——那是论证的核心。
Bounding $f$ from above only, with $f(x)\le h(x)$, is enough to apply the Squeeze Theorem. · 仅从上方对 $f$ 进行界定（使用 $f(x)\le h(x)$）就足以应用夹逼定理。
You need bounds on both · 两者 sides with equal limits — one bound proves nothing. · 你需要两侧都有界限且极限相等 — 仅有一个界限无法证明任何东西。
The Squeeze Theorem, with $\cos x\le\tfrac{\sin x}{x}\le 1$, gives $\displaystyle\lim_{x\to0}\tfrac{\sin x}{x}=$ · 夹逼定理结合 $\cos x\le\tfrac{\sin x}{x}\le 1$ 给出 $\displaystyle\lim_{x\to0}\tfrac{\sin x}{x}=$
As $x\to0$, $\cos x\to1$ and the upper bound is $1$, so the middle is squeezed to $1$. · 当 $x\to0$ 时，$\cos x\to1$ 和上界均趋向于 $1$，因此中间部分被挤压至 $1$。
If the lower bound heads to $2$ and the upper bound heads to $5$, the Squeeze Theorem tells you the trapped limit is... · 如果下界趋向于 $2$ 且上界趋向于 $5$，夹逼定理告诉你是...（被夹住的极限）
The bounds must have equal · 相等 limits; unequal walls give no conclusion. · 界限必须具有相等的极限；不等的墙壁无法得出结论。
The Squeeze Theorem needs the bounds to squeeze from both sides with equal limits. If you only bound $f$ from above ($f\le h$), or the two bounds head to different values, you have proved nothing. And the inequality must genuinely hold near $c$ — check it, don't assume it.
夹逼定理要求两个界从两侧用相等的极限来挤压。如果你只从上面界住 $f$($f\le h$),或两个界奔向不同值,你就什么都没证明。而且不等式必须在 $c$ 附近确实成立——要检查,别假设。
Show $\displaystyle\lim_{x\to0}x^2\cos\tfrac1x=0$.
- For all $x\neq0$: $-1\le\cos\tfrac1x\le1$.
- Multiply by $x^2\ (\ge0)$: $-x^2\le x^2\cos\tfrac1x\le x^2$.
- $\displaystyle\lim_{x\to0}(-x^2)=0$ and $\displaystyle\lim_{x\to0}x^2=0$.
- By the Squeeze Theorem, the middle limit is also $0$.
证明 $\displaystyle\lim_{x\to0}x^2\cos\tfrac1x=0$。
- 对所有 $x\neq0$:$-1\le\cos\tfrac1x\le1$。
- 乘以 $x^2\ (\ge0)$:$-x^2\le x^2\cos\tfrac1x\le x^2$。
- $\displaystyle\lim_{x\to0}(-x^2)=0$ 且 $\displaystyle\lim_{x\to0}x^2=0$。
- 由夹逼定理,中间的极限也是 $0$。
The Squeeze Theorem: if $g\le f\le h$ near $c$ and $\lim g=\lim h=L$, then $\lim f=L$. Use it when a function oscillates or otherwise resists direct methods — bound it above and below by functions with equal limits, verify the inequality near $c$, and $f$ is trapped onto $L$.
夹逼定理:若在 $c$ 附近 $g\le f\le h$ 且 $\lim g=\lim h=L$,则 $\lim f=L$。当函数振荡或以其他方式抵抗直接方法时使用它——用极限相等的函数从上下界住它,验证 $c$ 附近的不等式,$f$ 就被困在 $L$ 上。