Determining Limits Using Algebraic Properties of Limits · 利用极限的代数性质确定极限
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| limit laws/ˈlɪmɪt lɔːz/ | 极限法则 | jí xiàn fǎ zé |
| direct substitution/daɪˈrekt ˌsʌbstɪˈtjuːʃn/ | 直接代入 | zhí jiē dài rù |
| polynomial/ˌpɒlɪˈnəʊmɪəl/ | 多项式 | duō xiàng shì |
| rational/ˈræʃənl/ | 有理 | yǒu lǐ |
| indeterminate form/ˌɪndɪˈtɜːmɪnət fɔːm/ | 不定式 | bù dìng shì |
Limits obey friendly algebra
- Once you know the limits of two functions, you can combine them like ordinary numbers.
- Whatever $f$ and $g$ head toward, their sum heads toward the sum of those targets.
- The same works for differences, products, and (carefully) quotients.
- These are the limit laws 极限法则 — they let you break a messy limit into simple pieces.
极限遵守友好的代数
- 一旦知道两个函数的极限,你就能像普通数字一样把它们组合起来。
- 无论 $f$ 和 $g$ 各自奔向什么,它们的和就奔向那两个目标的和。
- 差、积、以及(小心处理的)商也同样成立。
- 这些就是极限法则——它们让你把一个杂乱的极限拆成简单的小块。
The limit laws, in one place
- Suppose $\displaystyle\lim_{x\to c} f(x)=L$ and $\displaystyle\lim_{x\to c} g(x)=M$. Then:
- Sum / difference: $\lim (f\pm g) = L\pm M$.
- Product: $\lim (f\cdot g) = L\cdot M$; constant multiple: $\lim (k f)=kL$.
- Quotient: $\displaystyle\lim \frac{f}{g}=\frac{L}{M}$, provided $M\neq 0$.
极限法则,一次列清
- 设 $\displaystyle\lim_{x\to c} f(x)=L$,$\displaystyle\lim_{x\to c} g(x)=M$。那么:
- 和 / 差: $\lim (f\pm g) = L\pm M$。
- 积: $\lim (f\cdot g) = L\cdot M$;常数倍: $\lim (k f)=kL$。
- 商: $\displaystyle\lim \frac{f}{g}=\frac{L}{M}$,前提是 $M\neq 0$。
Given $\lim f = 6$ and $\lim g = 3$, find $\lim \dfrac{f}{g}$. · 已知 $\lim f = 6$ 和 $\lim g = 3$，求 $\lim \dfrac{f}{g}$。
Quotient law (denominator limit $3\neq0$): $\frac{6}{3}=2$. · 商法则（分母极限 $3\neq0$）：$\frac{6}{3}=2$。
The quotient law $\lim\frac fg=\frac LM$ requires that $M$, the denominator's limit, is not ____. · 商法则 $\lim\frac fg=\frac LM$ 要求 $M$（分母的极限）不能是 ____。
If $M=0$ the law does not apply directly — you may have an indeterminate form. · 如果 $M=0$，则法则不能直接应用——你可能遇到了不定式。
Direct substitution: the fast lane
- For a polynomial 多项式 or a rational 有理 function, the laws add up to one shortcut: just plug $c$ in.
- $\displaystyle\lim_{x\to 3}(x^2-2x+1) = 3^2-2(3)+1 = 4$ — no table, no graph.
- Direct substitution 直接代入 works whenever the function is defined and continuous at $c$.
- This is always the first thing to try — it is fastest when it works.
直接代入:快车道
- 对一个多项式或有理函数,这些法则合起来就是一个捷径:直接把 $c$ 代进去。
- $\displaystyle\lim_{x\to 3}(x^2-2x+1) = 3^2-2(3)+1 = 4$——不用表格,不用图。
- 只要函数在 $c$ 处有定义且连续,直接代入就有效。
- 这永远是第一件要试的事——有效时它最快。
A polynomial has no surprises · 多项式没有意外
y = ax² + bx + c
Polynomials are smooth everywhere — drag the coefficients and see why plugging in $c$ (direct substitution) simply reads off the height. · 多项式处处光滑——拖动系数并理解为何直接代入 $c$（直接代入法）只需读出高度即可。
Evaluate · 评价 $\displaystyle\lim_{x\to 2}(3x^2 - x + 4)$. · 计算 $\displaystyle\lim_{x\to 2}(3x^2 - x + 4)$。
Direct substitution: $3(4)-2+4 = 14$. · 直接代入：$3(4)-2+4 = 14$。
For which limits does direct substitution give the answer immediately? · 对于哪些极限可以直接代入立即得到答案？
The third gives $\tfrac00$ (denominator limit $0$) so substitution stalls; the others give a finite value straight away. · 第三个给出 $\tfrac00$（分母极限 $0$），所以代入受阻；其他项直接给出有限值。
When substitution stalls
- Plug in and you might hit $\dfrac{0}{0}$: an indeterminate form 不定式.
- Indeterminate does not mean the limit fails — it means substitution alone can't decide, and more work is needed (next lesson).
- Beware the quotient law's fine print: it needs $M\neq0$. If the denominator's limit is $0$, the law does not apply directly.
- So: try substitution first; if it gives a real number, you are done; if it gives $\tfrac00$, switch strategies.
当代入卡住时
- 代入后你可能撞上 $\dfrac{0}{0}$:一个不定式。
- 不定式不意味着极限失败——它意味着仅靠代入无法判定,需要更多工作(下一课)。
- 当心商法则的附加条件:它要求 $M\neq0$。如果分母的极限是 $0$,该法则不能直接套用。
- 所以:先试代入;若得到一个真实数字,就完成了;若得到 $\tfrac00$,就换策略。
Direct substitution gives $\tfrac00$. This means... · 直接代入得到 $\tfrac00$。这意味着...
$\tfrac00$ is indeterminate: a cue to do algebra, not a final answer. · $\tfrac00$ 是不定式：这是进行代数运算的信号，而非最终答案。
When you meet a new limit, direct substitution is the sensible first thing to try. · 当你遇到新极限时，直接代入是合理的第一步尝试。
It is fastest when it works; only switch strategies if it yields an indeterminate form. · 当它有效时最快；只有当它产生不定式时才更换策略。
$\dfrac{0}{0}$ is indeterminate, not "zero" and not "undefined-so-DNE." It is a signal that the limit is hiding and you must simplify first. Do not report $\tfrac00$ as an answer, and do not conclude the limit does not exist — $\lim_{x\to2}\frac{x^2-4}{x-2}$ is $\tfrac00$ on substitution yet equals $4$.
$\dfrac{0}{0}$ 是不定式,既不是"零",也不是"没定义所以极限不存在"。它是一个信号:极限藏起来了,你必须先化简。不要把 $\tfrac00$ 当作答案,也不要断定极限不存在——$\lim_{x\to2}\frac{x^2-4}{x-2}$ 代入是 $\tfrac00$,却等于 $4$。
Find $\displaystyle\lim_{x\to 1}\dfrac{2x^2+3}{x+4}$.
- Numerator limit: $2(1)^2+3 = 5$. Denominator limit: $1+4 = 5$.
- The denominator's limit is $5\neq 0$, so the quotient law applies.
- $\displaystyle\lim_{x\to 1}\dfrac{2x^2+3}{x+4} = \dfrac{5}{5} = 1$ by direct substitution.
求 $\displaystyle\lim_{x\to 1}\dfrac{2x^2+3}{x+4}$。
- 分子极限:$2(1)^2+3 = 5$。分母极限:$1+4 = 5$。
- 分母极限是 $5\neq 0$,所以商法则适用。
- 由直接代入,$\displaystyle\lim_{x\to 1}\dfrac{2x^2+3}{x+4} = \dfrac{5}{5} = 1$。
The limit laws let you split a limit across sums, differences, products, and quotients (quotient needs a non-zero denominator limit). For polynomials and rationals this collapses to direct substitution — always try it first. A result of $\tfrac00$ is an indeterminate form: not an answer, but a cue to simplify.
极限法则让你把极限分拆到和、差、积、商上(商需要分母极限非零)。对多项式和有理函数,这归结为直接代入——永远先试它。结果为 $\tfrac00$ 是一个不定式:它不是答案,而是提示你去化简。