Removing Discontinuities · 消除不连续性
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| piecewise function/ˈpiːswaɪz ˈfʌŋkʃn/ | 分段函数 | fēn duàn hán shù |
Patching a hole
- A removable discontinuity is a hole where the limit exists but the value is wrong or missing.
- Since the limit is a perfectly good number, you can redefine the function at that one point to fill the hole.
- Set the point's value equal to the limit — the curve now flows through smoothly.
- This is repairing continuity: possible only for the removable type.
给洞打补丁
- 可去间断是一个洞:极限存在,但取值错误或缺失。
- 既然极限是一个很好的数,你就能在那一个点重新定义函数来填洞。
- 把该点的取值设为极限——曲线现在平滑地穿过。
- 这是在修复连续性:只对可去类型才可能。
The curve behind the hole · 空洞背后的曲线
y = x + 3
The cancelled function $y=x+3$ is a straight line — filling the hole just plots the point the line already passes through. · 约分后的函数 $y=x+3$ 是一条直线 — 填充空洞只是绘制了该直线已经经过的点。
Solving for a mystery constant
- Often a piecewise function 分段函数 has an unknown constant $k$, and you must pick $k$ to make it continuous.
- The rule is condition (3): make the limit equal the value at the seam.
- Set the two pieces equal at the join point and solve for $k$.
- One equation, one unknown — clean algebra.
求一个未知常数
- 常常一个分段函数里有未知常数 $k$,你必须选 $k$ 使它连续。
- 规则就是条件 (3):在接缝处让极限等于取值。
- 在接合点让两段相等并解出 $k$。
- 一个方程、一个未知数——干净的代数。
To make a piecewise function continuous at the seam, set the limit equal to the function ____. · 要使分段函数在接缝处连续，需将极限设为函数 ____。
That is continuity condition (3): limit $=$ value. · 这就是连续性条件(3)：极限 $=$ 值。
A worked pattern
- Suppose $f(x)=\begin{cases}x+k,&x;<2\\3x-1,&x;\ge2\end{cases}$ is to be continuous at $x=2$.
- Left limit: $2+k$. Right value: $3(2)-1=5$.
- Continuity needs $2+k=5$, so $k=3$.
- With $k=3$ the two pieces meet exactly — no jump, a continuous function.
一个范例模式
- 设 $f(x)=\begin{cases}x+k,&x;<2\\3x-1,&x;\ge2\end{cases}$ 要在 $x=2$ 处连续。
- 左极限:$2+k$。右取值:$3(2)-1=5$。
- 连续性需要 $2+k=5$,所以 $k=3$。
- 当 $k=3$,两段恰好相接——没有跳跃,一个连续函数。
For · 支持 $f(x)=\begin{cases}x+k,&x;<2\\3x-1,&x;\ge2\end{cases}$ continuous at $x=2$, find $k$. · 对于 $f(x)=\begin{cases}x+k,&x;<2\\3x-1,&x;\ge2\end{cases}$ 在 $x=2$ 处连续，求 $k$。
Set pieces equal at $2$: $2+k=5\Rightarrow k=3$. · 在 $2$ 处使各段相等：$2+k=5\Rightarrow k=3$。
Know what cannot be fixed
- Only removable holes are repairable — the limit must already exist.
- A jump (unequal one-sided limits) has a gap of real width: no single redefinition closes it.
- An infinite discontinuity blows up: no value tames an asymptote.
- So first classify; if the two-sided limit is missing, the discontinuity is permanent.
知道什么无法修复
- 只有可去的洞可修复——极限必须已经存在。
- 跳跃(两个单侧极限不等)有真实宽度的缺口:任何单点重新定义都合不上。
- 无穷间断会爆发:没有哪个取值能驯服渐近线。
- 所以先分类;若双侧极限缺失,间断就是永久的。
Which discontinuity can · can(能力与请求) be removed by redefining a single point? · 哪种不连续性可以通过重新定义单个点来消除？
Only the removable type has a two-sided limit to match; jumps and blow-ups cannot be fixed. · 只有可去类型具有双侧极限可供匹配；跳跃和爆炸无法修复。
Redefining $f(c)$ can create a two-sided limit that did not exist before. · 重新定义 $f(c)$ 可能会创建一个之前不存在的双侧极限。
A single point does not affect the limit; if the sides disagree, redefining changes nothing. · 单个点不影响极限；如果两侧不一致，重新定义毫无作用。
Select all · 所有 discontinuities that cannot be removed. · 选择所有不可去除的间断点。
Anything without a two-sided limit is permanent; only the removable hole can be filled. · 没有双侧极限的间断点是永久性的；只有可去孔洞可以被填充。
You can only "remove" a discontinuity where the two-sided limit already exists. If the left and right limits differ, setting $f(c)$ to any number still leaves the sides disagreeing — the function stays discontinuous. Redefining a point never creates a limit that wasn't there.
你只能"消除"双侧极限已经存在之处的间断。若左右极限不同,把 $f(c)$ 设成任何数,两侧仍然不一致——函数依旧不连续。重新定义一个点,永远造不出原本没有的极限。
To remove the hole of $\dfrac{x^2-25}{x-5}$ at $x=5$, define $f(5)$ equal to the limit. What value? · 要消除 $\dfrac{x^2-25}{x-5}$ 在 $x=5$ 处的空洞，定义 $f(5)$ 等于极限。该值是多少？
Limit $=\lim_{x\to5}(x+5)=10$, so $f(5)=10$. · 极限 $=\lim_{x\to5}(x+5)=10$，所以 $f(5)=10$。
Find $c$ so that $f(x)=\begin{cases}\dfrac{x^2-9}{x-3},&x;\neq3\\ c,&x;=3\end{cases}$ is continuous at $x=3$.
- The limit: $\displaystyle\lim_{x\to3}\dfrac{x^2-9}{x-3}=\lim_{x\to3}(x+3)=6$.
- Continuity requires $c=\lim_{x\to3}f(x)=6$.
- Setting $c=6$ fills the hole; $f$ is now continuous at $3$.
求 $c$ 使 $f(x)=\begin{cases}\dfrac{x^2-9}{x-3},&x;\neq3\\ c,&x;=3\end{cases}$ 在 $x=3$ 处连续。
- 极限:$\displaystyle\lim_{x\to3}\dfrac{x^2-9}{x-3}=\lim_{x\to3}(x+3)=6$。
- 连续性要求 $c=\lim_{x\to3}f(x)=6$。
- 设 $c=6$ 填上洞;$f$ 现在在 $3$ 处连续。
A removable discontinuity is repaired by redefining $f(c)$ to equal the (existing) limit. For a piecewise function with an unknown constant, set the pieces equal at the join — limit $=$ value — and solve. Jump and infinite discontinuities have no two-sided limit, so they cannot be removed.
可去间断通过把 $f(c)$ 重新定义为(已存在的)极限来修复。对含未知常数的分段函数,在接合点让两段相等——极限 $=$ 取值——并求解。跳跃与无穷间断没有双侧极限,所以无法消除。