导航菜单

极限法

利用极限的性质判定连续性。

判定原则

  • 如果 limxx0f(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)=sinxf(x) = \sin xx=0x = 0 处的连续性

  1. f(0)=sin0=0f(0) = \sin 0 = 0
  2. limx0sinx=0\lim_{x \to 0} \sin x = 0
  3. limx0f(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)=x21x1f(x) = \frac{x^2 - 1}{x - 1}x=1x = 1 处的连续性

  1. 函数在 x=1x = 1 处无定义
  2. limx1x21x1=limx1(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

练习题

练习 1

判断函数 f(x)={sinxx,x01,x=0f(x) = \begin{cases} \frac{\sin x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}x=0x = 0 处的连续性。

参考答案

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:limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1
  3. 比较:limx0f(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

    先修课程

    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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  2. 2

    Continuity in Advanced Calculus

    当前课程

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

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