极限法

利用极限的性质判定连续性。

判定原则

  • 如果 lim⁡x→x0f(x)=f(x0)\lim_{x \to x_0} f(x) = f(x_0),则函数在 x0x_0 点连续
  • 如果极限不存在或极限值不等于函数值,则函数在该点不连续

应用例子

例子 1:判断函数 f(x)=sin⁡xf(x) = \sin x 在 x=0x = 0 处的连续性

解:

  1. f(0)=sin⁡0=0f(0) = \sin 0 = 0
  2. lim⁡x→0sin⁡x=0\lim_{x \to 0} \sin x = 0
  3. lim⁡x→0f(x)=f(0)\lim_{x \to 0} f(x) = f(0)
  4. 结论:函数在 x=0x = 0 处连续

例子 2:判断函数 f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1} 在 x=1x = 1 处的连续性

解:

  1. 函数在 x=1x = 1 处无定义
  2. lim⁡x→1x2−1x−1=lim⁡x→1(x+1)=2\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} (x + 1) = 2
  3. 极限存在但函数值无定义
  4. 结论:函数在 x=1x = 1 处不连续

Exercises

练习 1

判断函数 f(x)={sin⁡xx,x≠01,x=0f(x) = \begin{cases} \frac{\sin x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases} 在 x=0x = 0 处的连续性。

Reference Answer

解题思路:使用极限法判定。

详细步骤:

  1. 函数在 x=0x = 0 处有定义:f(0)=1f(0) = 1
  2. 计算极限:lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1
  3. 比较:lim⁡x→0f(x)=f(0)=1\lim_{x \to 0} f(x) = f(0) = 1

答案:函数在 x=0x = 0 处连续。

课程路线图

  1. 1

    Exploring Functions in Advanced Mathematics

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    Functions are a core idea of advanced mathematics. This course walks through foundational concepts, key properties, and classic constants so you can read, reason, and compute with confidence.

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    Continuity in Advanced Calculus

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    A focused guide on continuity: core definitions, types of discontinuities, and continuity of elementary functions.

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