# 无穷极限与垂直渐近线的联系

> AP 微积分 BC · 第一单元：极限与连续性
> 来源: https://www.owlsprep.com/zh/study/ap-calculus-bc-u1-connecting-infinite-limits-and-vertical/

本小节将单侧无穷极限行为与垂直渐近线联系起来。你将学习如何确定有理函数、对数函数和三角函数的垂直渐近线位置，并正确区分渐近线与可去不连续点（洞）。

**先修:** 计算单侧极限; 因式分解与化简有理函数; 不连续性的定义

## 学习目标

- 定义单侧和双侧无穷极限
- 将无穷极限行为与垂直渐近线联系起来
- 确定有理函数、对数函数和三角函数的垂直渐近线位置
- 区分垂直渐近线与可去不连续点（洞）

## 无穷极限：定义与单侧行为

**无穷极限** — 描述当$x$趋近于常数$a$时，$f(x)$无界增长或衰减的记号。该极限不作为有限数存在；这个记号仅描述函数在$a$附近的行为。

*记法:* $\lim_{x \to a^\pm} f(x) = \pm\infty$

无穷极限几乎总是逐点单侧计算，因为结果的符号取决于你从$a$的哪一侧趋近。对于有理函数，如果分子趋近于非零常数，分母趋近于0，则结果为无穷极限，符号由$a$附近分子和分母的符号决定。

**例题:** Evaluate $\lim_{x \to 3^-} \frac{2x + 1}{x - 3}$ and $\lim_{x \to 3^+} \frac{2x + 1}{x - 3}$

1. 计算$x \to 3$时分子的极限：

   $$\lim_{x \to 3} (2x +1) = 2(3) + 1 = 7$$
2. 这是一个非零正常数。对于左极限（$x \to 3^-$），$x < 3$所以$x - 3 < 0$，从负侧趋近于0。正常数除以小负数得到大负数，因此：
3. $$\lim_{x \to 3^-} \frac{2x + 1}{x - 3} = -\infty$$
4. For the right-hand limit ($x \to 3^+$), $x > 3$ so $x - 3 > 0$, approaching 0 from the positive side. A positive constant divided by a small positive number is a large positive number, so:
5. $$\lim_{x \to 3^+} \frac{2x + 1}{x - 3} = +\infty$$

> **考试提示:** 检查无穷行为时，一定要分别计算单侧极限。AP考试经常考察你是否认识到双侧极限可能不存在（因为两侧趋向相反无穷），但单侧行为仍会产生垂直渐近线。

## 核心联系：由无穷极限得到垂直渐近线

**垂直渐近线** — 若当$x$趋近于$a$时，$f(x)$至少有一个单侧极限是无穷（$+\infty$或$-\infty$），则直线$x=a$是$f(x)$的垂直渐近线。这就是无穷极限和渐近线之间的核心形式联系。

*记法:* $x=a$ (a vertical line)

对于有理函数，这给出了寻找垂直渐近线的分步规则：(1) 完全因式分解分子和分母。(2) 约去公因子化简；公因子会产生可去不连续点（洞），而非渐近线。(3) 任何使化简后分母为零的$x=a$都是垂直渐近线。

**例题:** Find all vertical asymptotes of $f(x) = \frac{x^2 - 9}{x^2 - 2x - 3}$

1. 完全因式分解分子和分母：

   $$x^2 - 9 = (x-3)(x+3), \quad x^2 - 2x -3 = (x-3)(x+1)$$
2. Simplify, noting the domain restriction $x \neq 3$:

   $$f(x) = \frac{x+3}{x+1} \quad \text{for } x \neq 3$$
3. Find zeros of the simplified denominator: $x+1 = 0 \implies x = -1$. Check the one-sided limit at $x=-1$:

   $$\lim_{x \to -1^-} \frac{x+3}{x+1} = \frac{2}{0^-} = -\infty$$
4. This confirms an infinite one-sided limit, so $x=-1$ is a vertical asymptote. At $x=3$, evaluate the limit of the simplified function:

   $$\lim_{x \to 3} \frac{x+3}{x+1} = \frac{6}{4} = \frac{3}{2}$$
5. 极限是有限的，因此$x=3$是洞（可去不连续点），不是垂直渐近线。最终结果：只有$x=-1$是垂直渐近线。

> **考试提示:** AP考试的有理函数垂直渐近线问题几乎总会包含公因子，考察你是否会把洞和垂直渐近线混淆。识别渐近线前一定要先化简。

## 非有理函数的垂直渐近线

Rational functions are not the only functions with vertical asymptotes. For any function, find points where the function is undefined, then check if at least one one-sided limit at that point is infinite. Two common cases tested on the AP exam are:

- **对数函数**：对于$f(x) = \ln(g(x))$，$f$在$g(x) \leq 0$处无定义。垂直渐近线出现在$x=a$处，其中$g(x)$从正侧趋近于0（在$f$的定义域内）。
- **Trigonometric functions**: Reciprocal trigonometric functions like $\tan x$, $\cot x$, $\sec x$, and $\csc x$ have vertical asymptotes where their denominators are zero, since the numerator is non-zero at these points.

**例题:** Find all vertical asymptotes of $f(x) = \ln(x^2 - 4x + 3)$

1. First find the domain of $f(x)$: the argument of the logarithm must be positive:

   $$x^2 - 4x + 3 = (x-1)(x-3) > 0 \implies \text{Domain: } (-\infty, 1) \cup (3, \infty)$$
2. Candidates for vertical asymptotes are the domain boundaries: $x=1$ and $x=3$, where the argument equals zero.
3. Check $x=1$: as $x \to 1^-$ (from the domain side), $(x-1)(x-3)$ is positive and approaches 0. So:

   $$\lim_{x \to 1^-} \ln((x-1)(x-3)) = -\infty$$
4. Thus $x=1$ is a vertical asymptote. Check $x=3$: as $x \to 3^+$ (from the domain side), $(x-1)(x-3)$ is positive and approaches 0. So:

   $$\lim_{x \to 3^+} \ln((x-1)(x-3)) = -\infty$$
5. Thus $x=3$ is also a vertical asymptote.

> **考试提示:** For logarithmic functions, only check boundaries of the domain where the argument approaches 0 from the positive side. Points where the argument approaches 0 from the negative side are outside the domain, so no asymptote exists there.

## AP风格例题练习

**例题:** Which of the following gives all values of $x$ at which the graph of $f(x) = \frac{3x^2 - 10x - 8}{x(x^2 - 16)}$ has a vertical asymptote?<br>Options: (A) $x=4$ only, (B) $x=-4$ and $x=0$ only, (C) $x=-4$, $x=0$, and $x=4$, (D) $x=-\frac{2}{3}$ and $x=4$

1. 完全因式分解分子和分母：

   $$3x^2 - 10x -8 = (3x + 2)(x - 4), \quad x(x^2 -16) = x(x-4)(x+4)$$
2. Cancel the common $(x-4)$ factor, giving the simplified function:

   $$f(x) = \frac{3x+2}{x(x+4)} \quad \text{for } x \neq 4$$
3. The simplified denominator equals zero at $x=0$ and $x=-4$. One-sided limits at both points are infinite, so both are vertical asymptotes. At $x=4$, the limit is finite:

   $$\lim_{x \to 4} \frac{3x+2}{x(x+4)} = \frac{14}{32} = \frac{7}{16}$$
4. So $x=4$ is a hole, not an asymptote. The correct answer is (B).

**例题:** Let $f(x) = \frac{e^{2x}}{(x-2)(x+4)}$. (a) Find all candidate points for vertical asymptotes, and justify why each is a candidate. (b) Confirm whether each candidate is a vertical asymptote by evaluating one-sided limits. (c) State all vertical asymptotes and explain why there are no others.

1. Part (a): $f(x)$ is a quotient of continuous functions, so it is only undefined where the denominator equals zero. Set $(x-2)(x+4) = 0$, so candidates are $x=2$ and $x=-4$. These are candidates because $f$ is undefined at both points, so infinite limit behavior is possible.
2. Part (b): At $x=2$: $\lim_{x \to 2^-} \frac{e^{2x}}{(x-2)(x+4)} = \frac{e^4}{(\text{negative})(\text{positive})} = -\infty$, so $x=2$ is a vertical asymptote. At $x=-4$: $\lim_{x \to -4^-} \frac{e^{2x}}{(x-2)(x+4)} = \frac{e^{-8}}{(\text{negative})(\text{negative})} = +\infty$, so $x=-4$ is also a vertical asymptote.
3. Part (c): All vertical asymptotes are $x=2$ and $x=-4$. $f(x)$ is defined and continuous everywhere else on its domain, so there are no other points with possible infinite limit behavior, hence no additional vertical asymptotes.

## 常见错误

- **错误做法:** Claiming $x=2$ is a vertical asymptote of $f(x) = \frac{x^2 - 4}{x-2}$ because it makes the original denominator zero.
  - 原因: Students confuse undefined points with asymptotes, forgetting to check for common factors that create removable discontinuities.
  - 正确做法: Always simplify the function first, then check if the limit as $x$ approaches the undefined point is infinite; finite limits mean holes, not asymptotes.
- **错误做法:** Concluding $x=a$ is not a vertical asymptote because the two one-sided limits go to opposite infinities.
  - 原因: Students incorrectly believe both one-sided limits must go to the same infinity for an asymptote to exist.
  - 正确做法: Recall that any infinite one-sided limit (one or two sides) is enough to confirm a vertical asymptote at $x=a$, regardless of whether the two sides match.
- **错误做法:** Claiming $x=0$ is not a vertical asymptote of $f(x) = \frac{1}{x^2}$ because $\lim_{x \to 0} \frac{1}{x^2} = +\infty$ (so the limit does not exist as a finite number).
  - 原因: Students confuse 'the limit does not exist as a finite number' with 'no infinite behavior that creates an asymptote'.
  - 正确做法: Remember infinite limit notation describes unbounded behavior, not an existing finite limit; if $\lim_{x \to a} f(x) = \pm\infty$, $x=a$ is a vertical asymptote.
- **错误做法:** Stating that $x=0$ is a vertical asymptote of $f(x) = \frac{\sin x}{x}$ because the denominator is zero at $x=0$.
  - 原因: Students memorize 'denominator zero means vertical asymptote' without checking the limit.
  - 正确做法: Always evaluate the limit as $x$ approaches the undefined point; $\lim_{x \to 0} \frac{\sin x}{x} = 1$, so this is a removable discontinuity, not an asymptote.
- **错误做法:** For $f(x) = \ln(x-4)$, claiming $x=0$ is a vertical asymptote because $\ln(0)$ is undefined.
  - 原因: Students forget to check what input makes the argument of the logarithm zero.
  - 正确做法: For logarithmic functions, set the argument equal to zero to find the candidate vertical asymptote, then confirm the limit is infinite from the domain side.

## 速查表

| 类别 | 公式 / 规则 | 注释 |
| --- | --- | --- |
| 右无穷极限 | $\lim_{x \to a^+} f(x) = \infty$ | $f(x)$ grows without bound as $x$ approaches $a$ from the right; limit does not exist as a finite number |
| 左负无穷极限 | $\lim_{x \to a^-} f(x) = -\infty$ | $f(x)$ decreases without bound as $x$ approaches $a$ from the left |
| 垂直渐近线定义 | $x=a$ is a VA if at least one $\lim_{x \to a^\pm} f(x) = \pm\infty$ | Only requires one infinite one-sided limit; both sides do not need to match |
| Rational function VAs | After canceling common factors, $x=a$ is a VA if simplified denominator $= 0$ at $x=a$ | Canceled common factors create holes (removable discontinuities), not VAs |
| Logarithm VAs | $y = \ln(g(x))$ has VA at $x=a$ if $\lim_{x \to a^\pm} g(x) = 0^+$ | Only check boundaries of the domain of $f$ |
| Tangent VAs | $y = \tan x$ has VAs at $x = \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$ | Follows from $\tan x = \frac{\sin x}{\cos x}$; VAs where $\cos x = 0$ |
| Hole vs Vertical Asymptote | If $\lim_{x \to a} f(x)$ is finite, $x=a$ is a hole; if infinite, it is a VA | All holes in rational functions come from common factors |

## 下一步

本主题是第一单元《极限与连续性》其余内容的核心基础，也是AP微积分BC课程后续核心主题的预备知识。接下来，你将学习无穷远处的极限与水平渐近线的联系，然后利用对渐近线的理解完整绘制函数曲线，包括一阶导数和二阶导数的图形。本主题对BC专属的反常积分主题也至关重要，在反常积分中你必须识别垂直渐近线，才能正确分类和计算含不连续点函数的反常积分。如果不能从无穷极限正确识别垂直渐近线，你会错误分类积分的不连续点，在FRQ试题中失分。

- [反常积分](https://www.owlsprep.com/zh/study/ap-calculus-bc-u6-improper-integrals/)
- [无穷远处极限与水平渐近线的联系](https://www.owlsprep.com/zh/study/ap-calculus-bc-u1-connecting-limits-at-infinity-and/)
- [介值定理（IVT）应用](https://www.owlsprep.com/zh/study/ap-calculus-bc-u1-working-with-the-intermediate-value/)

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