无穷极限与垂直渐近线的联系

AP 微积分 BC· AP Calculus BC CED — 极限与连续性· 14 分钟阅读

1. 无穷极限：定义与单侧行为★★☆☆☆⏱ 4 min

📘 定义

无穷极限

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

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。

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 $lim_{x \to 3^{-}} \frac{2 x + 1}{x - 3}$ and $lim_{x \to 3^{+}} \frac{2 x + 1}{x - 3}$

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
$\lim_{x \to 3^+} \frac{2x + 1}{x - 3} = +\infty$

Exam tip:

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。

2. 核心联系：由无穷极限得到垂直渐近线★★☆☆☆⏱ 4 min

📘 定义

垂直渐近线

$x = a$ (a vertical line)

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。

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 $f (x) = \frac{x ^{2} - 9}{x ^{2} - 2 x - 3}$

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: $x + 1 = 0 ⟹ x = - 1$ . 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。

Exam tip:

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。

3. 非有理函数的垂直渐近线★★★☆☆⏱ 4 min

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 $f (x) = ln (x^{2} - 4 x + 3)$

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.

Exam tip:

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.

4. AP风格例题练习★★★☆☆⏱ 6 min

📐 例题

Which of the following gives all values of $x$ at which the graph of $f (x) = \frac{3 x ^{2} - 10 x - 8}{x ( x ^{2} - 16 )}$ has a vertical asymptote?
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 $f (x) = \frac{e ^{2 x}}{( x - 2 ) ( x + 4 )}$ . (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 $x = 2$ : $lim_{x \to 2^{-}} \frac{e ^{2 x}}{( x - 2 ) ( x + 4 )} = \frac{e ^{4}}{( negative ) ( positive )} = - \infty$ , so $x = 2$ is a vertical asymptote. At $x = - 4$ : $lim_{x \to - 4^{-}} \frac{e ^{2 x}}{( x - 2 ) ( x + 4 )} = \frac{e ^{- 8}}{( negative ) ( negative )} = + \infty$ , 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.

5. 常见陷阱

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 $f (x) = \frac{s i n x}{x}$ 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; $lim_{x \to 0} \frac{s i n x}{x} = 1$ , 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.

6. 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 $x = \frac{π}{2} + k π, k \in Z$	Follows from $tan x = \frac{s i n x}{c o s x}$ ; 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

真题中的出现

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2023 · MCQ
求有理函数的垂直渐近线
2022 · FRQ
证明垂直渐近线的存在

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。