Limits & Continuity
AP Calculus BC Topic 1 17:56 English narration · English + 中文 subtitles burned in
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Transcript
Start with a puzzle.
先看一个谜题。
A car's speedometer reads sixty kilometres per hour.
汽车的速度表显示每小时六十公里。
But what does that mean at a single instant?
但在某一瞬间，这是什么意思呢？
Speed is distance over time.
速度是距离除以时间。
Yet in one instant, no time passes and no distance is covered — so the fraction becomes zero divided by zero, which is undefined.
可是在一个瞬间里，没有时间流逝，也没有走过距离—— 于是这个分数变成零除以零，没有定义。
Calculus escapes this trap with one idea: the limit.
微积分用一个思想跳出了这个陷阱：极限。
This unit builds the foundation of the whole course: limits, and the continuity they define.
本单元要打下整门课程的基础：极限，以及由它定义的连续性。
Let's begin.
让我们开始吧。
The trick is not to force the forbidden division. Instead, we watch where it is heading.
诀窍不是去硬做那个被禁止的除法，而是去观察它趋向何处。
Slide the input toward the point, from both sides at once, and follow the output.
让输入从两侧同时滑向那个点，并跟随输出。
If both sides aim at the same height, that height is the limit — even when the function never actually reaches it.
如果两侧瞄准同一个高度， 那个高度就是极限——即使函数从未真正到达它。
The open circle marks the value the curve is aiming for.
空心圆标出曲线所瞄准的值。
Calculus is the mathematics of change and of accumulation, and it answers two big questions: how fast is something changing right now, and how much has piled up so far.
微积分是研究变化和累积的数学，它回答两个大问题：某样东西此刻变化得有多快， 以及到目前为止累积了多少。
Both rest on one tool — the limit.
两者都依赖同一个工具——极限。
Here is the distinction that sets it up.
下面这个区分为它做好了铺垫。
The average rate of change uses a whole interval: the change in one quantity divided by the change in another.
平均变化率用的是一整个区间： 一个量的变化除以另一个量的变化。
Shrink the interval to nothing and it divides by zero, so it is undefined.
把区间缩到零，它就变成除以零，于是没有定义。
The instantaneous rate of change is what we actually want, at a point.
瞬时变化率才是我们真正想要的，是在某一点上的。
And the clever move is not to plug in zero, which is undefined, but to watch what the average rate APPROACHES as the interval shrinks.
而巧妙之处不在于代入零——那没有定义—— 而在于观察当区间不断缩小时，平均变化率趋近于什么。
That approaching value is a limit — and it powers the derivative, and run in reverse, the integral.
那个趋近的值就是极限—— 它支撑起导数，反过来运行，又支撑起积分。
A limit can be shown three ways, and the AP exam expects all three.
极限可以用三种方式呈现，而 AP 考试三种都要求。
Graphically: trace the curve toward the point and read the height it heads for.
用图象：沿着曲线走向那一点， 读出它趋向的高度。
Numerically: build a table of inputs creeping toward the point from both sides and watch the outputs.
用数值：列一张表，让自变量从两侧不断逼近那一点，观察函数值。
Analytically: use algebra to get an exact value.
用解析式：用代数求出精确值。
Moving between them is itself a core skill — a graph shows shape and any holes, a table gives numerical evidence, and algebra gives the exact answer and the reason.
在三者之间转换本身就是一项核心技能—— 图象显示形状以及有没有空洞，表格给出数值证据，代数给出精确答案和理由。
Strong answers use one representation to CONFIRM another.
好的答案会用一种表示去印证另一种。
And one note: the epsilon-delta definition of a limit is not tested on the AP exam, so you do not need it.
还有一点： 极限的 epsilon-delta 定义不在 AP 考试范围内，所以你不需要它。
We write it like this, and read it out loud: The limit of f of x, as x approaches c, equals L.
我们这样书写，并且这样读出来：当 x 趋近 c 时，f x 的极限等于 L。
Listen to those words — as x approaches c, <slow>not equals c.</slow> A limit means f of x can be made arbitrarily close to L by taking x sufficiently close to c — the behaviour of a function near a point, never the value at the point itself.
注意这几个字——是 x 趋近 c，不是等于 c。 极限描述的是函数在一点附近的行为， 绝不是它在那一点本身的值。
The function may be undefined there, or defined but different.
函数在那里可能没有定义，或者有定义但取别的值。
The limit does not care.
极限并不在乎。
A graph is often the fastest way to read a limit.
看图往往是读极限最快的方法。
Trace in from the left, then from the right, and see where each side is heading.
先从左边趋近，再从右边趋近，看看每一侧奔向何处。
Here both sides agree on the height three, so the limit is three.
这里两侧都认同高度三，所以极限是三。
But the function's actual value — the filled dot — sits elsewhere, up at five.
但函数的真实取值——那个实心点——却在别处， 在五那么高。
The open circle is what the curve approaches; the filled dot is what the function equals.
空心圆是曲线所趋近的值；实心点是函数所等于的值。
They need not match.
两者不必一致。
This picture is the single most important idea in the unit.
这张图是本单元最重要的一个想法。
A limit describes the behaviour of f NEAR c, not the value AT c.
极限描述的是 f 在 c 附近的行为， 而不是它在 c 处的值。
The function may be undefined at c, or defined but equal to something else entirely — the limit does not care.
函数在 c 处可能没有定义，也可能有定义却等于完全不同的东西—— 极限不在乎。
Graphs mark the difference: an open circle marks a height the curve approaches but does not reach, a hole; a closed circle marks the actual value f of c.
图象把这个区别标了出来：空心圆表示曲线趋近却没有达到的高度， 也就是一个空洞；实心圆表示实际的函数值 f 在 c 处的值。
So a curve can approach three from both sides, making the limit three, while a filled dot sits at height five, making f of c equal five.
所以曲线可以从两侧都趋近于三，于是极限是三， 而一个实心点却落在高度五上，于是 f 在 c 处的值是五。
The limit is three.
极限是三。
The two need not match, and a question will test exactly that.
两者不必相同，而考题考的正是这一点。
Watch the scale of a graph too: a zoomed-out picture can hide important behaviour near a point, so confirm with algebra when you can.
还要注意图象的比例： 缩小的图可能掩盖一点附近的重要行为，所以能用代数确认时就确认一下。
When you have data or a formula but no picture and no easy algebra, a table estimates a limit numerically.
当你手上有数据或者公式，却没有图象、代数也不好做时，可以用表格来数值地估计极限。
Build it by choosing inputs that creep toward c from both sides.
做法是选取从两侧不断逼近 c 的自变量值。
Take the limit as x approaches two of x squared minus four, over x minus two — which is zero over zero at x equals two.
以 x 趋于二时 x 平方减四除以 x 减二为例—— 在 x 等于二处它是零比零。
From the left: one point nine gives three point nine, one point nine nine gives three point nine nine, one point nine nine nine gives three point nine nine nine.
从左边：一点九给出三点九，一点九九给出三点九九， 一点九九九给出三点九九九。
From the right: two point zero zero one gives four point zero zero one, and so on.
从右边：二点零零一给出四点零零一，依此类推。
Both sides march toward four, so we estimate the limit is four.
两侧都朝着四前进，所以我们估计极限是四。
But note the caution: a table only SUGGESTS a value.
但要注意：表格只是暗示一个值。
It is numerical evidence, not a proof.
它是数值证据，不是证明。
Tracing from the left gives the left-hand limit — inputs smaller than c.
从左边趋近给出左极限——自变量小于 c。
Tracing from the right gives the right-hand limit — inputs larger than c.
从右边趋近给出右极限——自变量大于 c。
These are the one-sided limits.
这些就是单侧极限。
If both head to the same height R, then the two-sided limit exists and equals R.
如果两侧都奔向同一个高度 R，那么双侧极限存在且等于 R。
If they disagree, the two-sided limit does not exist — DNE — even when each one-sided limit is finite.
如果它们不一致，双侧极限就不存在，即便每个单侧极限都是有限的。
Crucially, ignore the point itself when you read either side.
关键的是：读任何一侧时都要忽略那个点本身。
Sometimes no limit exists at all.
有时极限根本不存在。
Three classic ways.
有三种经典情形。
First, a jump: the left and right sides head to different heights, so they disagree.
第一，跳跃：左侧和右侧奔向不同的高度，彼此不一致。
Second, it grows without bound, running away toward infinity.
第二，它无界增长，一路跑向无穷。
Third, it oscillates, wobbling faster and faster and never settling.
第三，它振荡，摆动得越来越快，始终无法安定。
In each case we say the limit does not exist.
这三种情形我们都说极限不存在。
Classic examples: one over x squared as x approaches zero goes to infinity; the absolute value of x over x jumps at zero; and sine of one over x oscillates forever.
经典例子：当 x 趋于零时，一除以 x 平方趋于无穷； x 的绝对值除以 x 在零处跳跃；而一除以 x 的正弦则永远振荡。
To compute a limit by hand, always try direct substitution first — just put the value in.
要手算一个极限，永远先试直接代入——把值代进去。
A real number means you are done.
得到一个实数，就完成了。
But sometimes you get zero over zero: an indeterminate form.
但有时你会得到零比零：这是未定式。
That does not mean the limit fails; it means rewrite.
它并不意味着极限失败，而是意味着要改写。
Factor and cancel the common piece, then substitute again.
把公因式分解并约去，再代入一次。
Here the limit is four — and the factor you cancelled is exactly why the graph has a hole there.
这里极限是四—— 而你约去的那个因式，正是图象在那里有一个空洞的原因。
Most limits are found analytically using the limit laws.
大多数极限是用极限定理解析地求出来的。
The idea in one sentence: if both limits exist, the limit of a combination is the same combination of the limits.
一句话概括这个想法： 如果两个极限都存在，那么组合的极限就等于极限的同样组合。
So for sum, difference and product, just take the limits separately and combine them.
所以对于和、差、积，只要分别求极限再组合起来就行。
The quotient rule has a condition attached — the limit of a quotient is the quotient of the limits, PROVIDED the bottom limit is not zero, and that proviso is where most of this unit's difficulty lives.
商的法则附带一个条件——商的极限等于极限的商，前提是分母的极限不为零， 而本单元的大部分难点，正是藏在这个前提里。
The composite rule: if g is continuous at the limit of f, you can move the limit inside g.
复合函数法则： 如果 g 在 f 的极限处连续，你就可以把极限移到 g 里面去。
And the practical consequence: for a function built from polynomials and roots, try direct substitution first.
而实际的推论是：对于由多项式和根式搭起来的函数，先试直接代入。
If you get a real number, that is the limit and you are done.
如果得到一个实数，那就是极限，你就做完了。
Direct substitution sometimes gives zero over zero — the indeterminate form.
直接代入有时会给出零比零——也就是未定式。
And here is the key reading: that does NOT mean the limit fails.
而关键的理解在于： 这并不意味着极限不存在。
It means you must rewrite the function into an equivalent form that removes the trouble — use alternate forms — then substitute.
它意味着你必须把函数改写成一个等价的、 去掉了麻烦的形式，然后再代入。
Three standard moves.
有三个标准手法。
Factor and cancel, for a rational function: x squared minus four over x minus two becomes x plus two, and substituting gives four.
因式分解并约分，用于有理函数： x 平方减四除以 x 减二化为 x 加二，代入后得到四。
Multiply by the conjugate, for a radical: root of x plus one, minus one, all over x, becomes one over root of x plus one, plus one — which is a half.
乘以共轭式，用于根式： 根号 x 加一再减一，除以 x，化为一除以根号 x 加一再加一——结果是二分之一。
And use trigonometric identities to simplify — use alternate forms of the trig functions.
以及用三角恒等式来化简。
One last insight: the cancelled factor is exactly WHY the original graph had a hole.
最后还有一个洞见：被约掉的那个因式， 恰恰就是原来的图象上有一个空洞的原因。
The two functions agree everywhere except at c, so they share the same limit there.
这两个函数在除 c 以外的所有地方都相同， 所以它们在那里有相同的极限。
Put those together into a decision list, because on the exam the hard part is choosing.
把这些整理成一张判断清单，因为在考试中，难的部分是做选择。
One: try direct substitution first — a real answer means you are done.
第一：先试直接代入——得到一个实数就做完了。
Two: getting zero over zero?
第二：得到零比零？
Rewrite — factor and cancel, or the conjugate, or a trig identity — then substitute.
就改写——因式分解并约分，或者用共轭式，或者用三角恒等式——然后代入。
Three: a non-zero number over zero, like five over zero?
第三：非零数除以零，比如五除以零？
That limit is infinite or does not exist, and you must check the sign from each side.
那个极限是无穷大或者不存在， 而且你必须分别检查两侧的符号。
Four: as x goes to infinity?
第四：x 趋于无穷？
Compare the dominant terms.
比较主导项。
Five: trapped between two functions?
第五：被夹在两个函数之间？
The squeeze theorem may apply.
那可能要用夹逼定理。
Run down that list in order and you will pick the right tool nearly every time.
照这个顺序走一遍，你几乎每次都能选对工具。
Some functions are too wild to handle directly.
有些函数太狂野，无法直接处理。
Take x squared, times the sine of one over x.
看 x 平方 乘以 x 分之一 的正弦。
Near zero it oscillates infinitely fast — no substitution will ever work.
在零附近它以无限快的速度振荡——任何代入都不管用。
But we can trap it.
但我们可以把它夹住。
Because sine never leaves minus one and one, the whole thing is squeezed between minus x squared and x squared.
因为正弦永远不超出负一到一，整个式子就被夹在 负 x 平方 和 x 平方 之间。
Both of those bounds go to zero — so the wild function is forced to zero as well.
这两个界都趋于零——所以那个狂野的函数也被迫趋于零。
The squeeze theorem also proves two of the most useful limits in calculus.
夹逼定理还证明了微积分中最有用的两个极限。
Sine of x over x approaches one as x approaches zero.
当 x 趋近零时，x 的正弦除以 x 趋近一。
And one minus cosine of x, over x, approaches zero.
而一减去 x 的余弦，再除以 x，趋近零。
Both are used everywhere later — derivatives of sine and cosine rest on them.
这两个结果后面处处会用到—— 正弦和余弦的导数就建立在它们之上。
Remember the pattern: if you can trap a function between two others that share a limit, the trapped function is forced to the same place.
记住这个模式： 如果你能把一个函数夹在另外两个共享同一极限的函数之间，被夹住的函数就被迫去同一个地方。
Continuity is defined by three conditions, and all three must hold.
连续性由三个条件定义，而且三个必须同时成立。
One: the function has a value at the point.
第一：函数在该点有值。
Two: the limit exists there.
第二：极限在那里存在。
Three: the two are equal.
第三：二者相等。
In plain words — the point is there, the limit is there, and they agree.
用大白话说——点在那里，极限在那里， 而且它们一致。
If any one fails, the function is discontinuous.
只要有一条不成立，函数就在该点间断。
Treat it as a checklist and you will never lose these marks.
把它当成一张核对清单， 你就永远不会丢掉这些分。
A function is continuous on an interval if it is continuous at every point of that interval.
如果一个函数在某个区间的每一点都连续，就说它在这个区间上连续。
But you rarely check point by point, because whole families are continuous on their domains: polynomial, rational, power, exponential, logarithmic and trigonometric functions.
但你很少需要逐点检查，因为整整几族函数在它们的定义域上都是连续的： 多项式函数、有理函数、幂函数、指数函数、对数函数和三角函数。
Note the words "on their domains" — that is doing the work.
注意"在它们的定义域上"这几个字——它们是关键。
So a rational function is continuous everywhere EXCEPT where its denominator is zero.
所以有理函数处处连续， 除了分母为零的地方。
The natural log is continuous for x greater than zero.
自然对数在 x 大于零时连续。
Knowing this lets you declare continuity quickly and correctly, which matters enormously, because the Intermediate Value Theorem question later needs exactly that statement.
掌握这一点能让你迅速而准确地断言连续性，而这非常重要， 因为后面介值定理的题目要的正是这样一句话。
If the limit exists at a hole, the discontinuity is removable — and the word means what it says.
如果在一个空洞处极限存在，那么这个间断就是可去的——这个词就是字面意思。
Redefine the function at that one point to equal the limit, and the graph is repaired.
把函数在那一点重新定义为等于该极限，图象就补好了。
Formally, set the missing value to the limit as x approaches c.
形式上说，就是把缺失的那个值设为 x 趋于 c 时的极限。
Now the exam version of this.
现在看它的考试版本。
For a piecewise-defined function, continuity at a boundary needs the two pieces to MEET: the left piece's value, the right piece's value, and f of c must all be equal.
对于分段函数，在分界点处连续，要求两段能够接上： 左边那一段的值、右边那一段的值，以及 f 在 c 处的值，三者必须全部相等。
That is a very common setup, and what it asks you to do is solve for a parameter — an unknown constant chosen to make the pieces match.
这是一种非常常见的题型，而它要你做的，是解出一个参数—— 一个待定常数，取值使得两段能接上。
Set the left-hand limit equal to the right-hand limit equal to f of c, and solve.
令左极限等于右极限等于 f 在 c 处的值，然后求解。
When continuity fails, the graph breaks in one of three ways.
当连续性失败时，图象会以三种方式之一断裂。
A removable discontinuity is a single hole — the limit exists, but the point is missing or misplaced, so you could patch it.
可去间断是一个单独的空洞—— 极限存在，但那个点缺失或放错了位置，所以你可以把它补上。
A jump discontinuity is where the two sides disagree and the curve steps.
跳跃间断是两侧不一致，曲线出现台阶。
An infinite discontinuity is where the function blows up at a vertical asymptote.
无穷间断是函数在垂直渐近线处发散。
Removable, jump, infinite — name all three.
可去、跳跃、无穷——三种都要叫得出名字。
Hold this picture in your head for the exam.
考试时把这张图记在脑子里。
Removable: a hole, limit exists, patchable by redefining one point.
可去：一个空洞，极限存在，改定义一个点就能补上。
Jump: left and right limits both exist but disagree, so the curve steps.
跳跃：左右极限都存在但不一致，曲线出现台阶。
Infinite: the function blows up at a vertical asymptote.
无穷：函数在垂直渐近线处发散。
When a continuity question asks you to classify the break, these are the only three answers.
当连续性题目让你给断裂分类时，答案只有这三种。
Two kinds of asymptote come straight out of limits.
两类渐近线都直接来自极限。
A vertical asymptote appears where a non-zero number is divided by something shrinking to zero — the graph hugs a vertical line and runs off toward infinity.
垂直渐近线出现在一个非零的数被某个趋于零的量所除的地方—— 图象紧贴一条竖直线，冲向无穷。
Check each side, because the two sides can run opposite ways.
要分别检查两侧，因为两侧可能奔向相反的方向。
A horizontal asymptote comes from letting the input grow instead.
水平渐近线则来自让输入不断增大。
If the outputs settle toward a finite height, that line records the end behaviour.
如果输出稳定到一个有限的高度， 那条线就记录了函数的末端行为。
When a function grows without bound near x equals c, we write the limit as plus or minus infinity — an infinite limit.
当函数在 x 等于 c 附近无限增大时，我们把极限写成正无穷或负无穷——这就是无穷极限。
That describes a vertical asymptote: the graph hugs the vertical line and shoots off.
它描述的是一条垂直渐近线：图象紧贴着这条竖直线冲出去。
It happens where a NON-ZERO number is divided by something approaching zero, typically at a zero of the denominator that does not cancel.
它发生在一个非零数除以某个趋于零的东西的地方， 通常是在分母的某个不能约掉的零点上。
Note that qualifier — if it cancels, you get a hole instead, not an asymptote.
注意那个限定语—— 如果能约掉，你得到的就是一个空洞，而不是渐近线。
And always check each side separately, because the two sides can shoot OPPOSITE ways.
而且一定要分别检查两侧，因为两侧可能冲向相反的方向。
In this picture the left side falls to minus infinity and the right side climbs to plus infinity, so the two-sided limit does not exist.
在这张图里，左侧落向负无穷，右侧升向正无穷，所以双侧极限不存在。
We can also let the INPUT grow.
我们也可以让自变量增大。
Limits at infinity describe the end behaviour of a function as x goes to plus or minus infinity, and if the outputs settle toward a finite value L, then y equals L is a horizontal asymptote.
无穷远处的极限描述的是当 x 趋于正无穷或负无穷时， 函数的末端行为；如果函数值稳定趋向一个有限值 L，那么 y 等于 L 就是一条水平渐近线。
For a rational function there is a three-case rule you should know cold.
对有理函数有一条三种情形的规则，你应当烂熟于心。
If the top degree is LESS than the bottom degree, the limit is zero, so the asymptote is y equals zero.
如果分子的次数小于分母的次数，极限是零，所以渐近线是 y 等于零。
If the degrees are EQUAL, the limit is the ratio of the leading coefficients.
如果两者次数相等，极限就是首项系数之比。
And if the top degree is GREATER, the function is unbounded and there is no horizontal asymptote.
如果分子的次数更大，函数就无界，没有水平渐近线。
More generally you are comparing relative magnitudes — relative growth rates: far out, an exponential beats any polynomial, and a polynomial beats any logarithm.
更一般地说，你是在比较相对的增长速度：在很远处， 指数函数快过任何多项式，而多项式又快过任何对数函数。
Let's work one.
我们来做一道。
Find the limit of three x squared minus five, over two x squared plus x, as x runs off to infinity.
求当 x 趋于无穷时，三 x 平方 减五，除以 二 x 平方 加 x 的极限。
The standard move: divide the top and the bottom by the highest power, x squared.
标准做法是：把分子和分母都除以最高次幂，也就是 x 平方。
That leaves three minus five over x squared, over two plus one over x.
于是剩下 三 减 x 平方分之五，除以 二 加 x 分之一。
As x grows huge, those small pieces vanish — they go to zero.
当 x 变得很大时， 那些小项就消失了——它们趋于零。
What remains is three over two.
剩下的就是二分之三。
And because the limit is a finite number, the line y equals three halves is a horizontal asymptote of the graph.
而因为极限是一个有限的数，直线 y 等于二分之三 就是图象的一条水平渐近线。
Finally, a guarantee.
最后是一个保证。
The Intermediate Value Theorem — the IVT — says: if a function is continuous on a closed interval, it takes every height between its two endpoint values.
介值定理说：如果一个函数在闭区间上连续， 它就会取到两个端点值之间的每一个高度。
An unbroken curve cannot skip a height — it has to pass through.
一条不间断的曲线不能跳过任何高度——它必须经过。
That is how we guarantee that a root exists: negative at one end, positive at the other, so it must cross zero somewhere between.
这正是我们保证根存在的方法：一端为负，另一端为正，那么它必定在中间某处穿过零。
To justify it for full marks you must say three things — the function is continuous, the target lies between the endpoints, and therefore by the theorem the value is reached.
要拿满分，你必须说三句话——函数连续，目标值落在两端点之间， 因此由该定理，这个值一定被取到。
Intermediate Value Theorem questions appear almost every year — "must there be a value c with R of c equals a hundred and fifty-five?", or "is there a time when r-prime of t equals minus six?"
介值定理的题目几乎每年都会出现——"是否一定存在某个 c 使得 R 在 c 处的值等于一百五十五？ " 或者"是否存在某个时刻使得 r 撇 t 等于负六？
A full-credit justification has exactly three moves, and you should write all three every time.
"一份能拿满分的论证恰好有三步， 而且每一次三步都要写出来。
One: state continuity — say the function is continuous on the closed interval, often because it is differentiable or because you were told it is continuous.
第一：陈述连续性——说明这个函数在闭区间上连续， 通常是因为它可导，或者因为题目已经告诉你它连续。
Two: show d is trapped — compute the two endpoint values and show your target lies BETWEEN them.
第二：说明 d 被夹住了——算出两个端点的值，并说明你的目标值落在它们之间。
Three: conclude by name — "by the Intermediate Value Theorem — an existence theorem — there is at least one c in the open interval with f of c equals d."
第三：点名下结论——"根据介值定理，在开区间内存在一个 c 使得 f 在 c 处的值等于 d。
Skipping either one of the first two loses the point, because the theorem REQUIRES both conditions.
" 前两步中漏掉任何一步都会丢分，因为这个定理要求两个条件同时成立。
Before you go, three marks students often lose.
结束之前，三个学生常失的分。
First, a limit is what the function approaches — it need not equal the function's value, and both sides must agree.
第一，极限是函数所趋近的值——它不一定等于函数值， 而且两侧必须一致。
Second, when substitution gives zero over zero, do not stop: factor and cancel, or divide by the highest power, then substitute.
第二，当代入给出零比零时，不要停下：因式分解并约分， 或者除以最高次幂，然后再代入。
Third, to use the Intermediate Value Theorem you must state that the function is continuous and that your target lies between the endpoints.
第三，要用介值定理，你必须说明函数连续， 并且你的目标值落在两端点之间。
Get these right, and this topic is yours.
把这些做对，这个专题就是你的了。