极限法
利用极限的性质判定连续性。
判定原则
如果 lim x → x 0 f ( x ) = f ( x 0 ) \lim_{x \to x_0} f(x) = f(x_0) lim x → x 0 f ( x ) = f ( x 0 ) ,则函数在 x 0 x_0 x 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 x f(x) = \sin x f ( x ) = sin x 在 x = 0 x = 0 x = 0 处的连续性
解 :
f ( 0 ) = sin 0 = 0 f(0) = \sin 0 = 0 f ( 0 ) = sin 0 = 0
lim x → 0 sin x = 0 \lim_{x \to 0} \sin x = 0 lim x → 0 sin x = 0
lim x → 0 f ( x ) = f ( 0 ) \lim_{x \to 0} f(x) = f(0) lim x → 0 f ( x ) = f ( 0 )
结论:函数在 x = 0 x = 0 x = 0 处连续
例子 2 :判断函数 f ( x ) = x 2 − 1 x − 1 f(x) = \frac{x^2 - 1}{x - 1} f ( x ) = x − 1 x 2 − 1 在 x = 1 x = 1 x = 1 处的连续性
解 :
函数在 x = 1 x = 1 x = 1 处无定义
lim x → 1 x 2 − 1 x − 1 = lim x → 1 ( x + 1 ) = 2 \lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} (x + 1) = 2 lim x → 1 x − 1 x 2 − 1 = lim x → 1 ( x + 1 ) = 2
极限存在但函数值无定义
结论:函数在 x = 1 x = 1 x = 1 处不连续
练习题
练习 1
判断函数 f ( x ) = { sin x x , x ≠ 0 1 , x = 0 f(x) = \begin{cases} \frac{\sin x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases} f ( x ) = { x s i n x , 1 , x = 0 x = 0 在 x = 0 x = 0 x = 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 :
函数在 x = 0 x = 0 x = 0 处有定义:f ( 0 ) = 1 f(0) = 1 f ( 0 ) = 1
计算极限:lim x → 0 sin x x = 1 \lim_{x \to 0} \frac{\sin x}{x} = 1 lim x → 0 x s i n x = 1
比较:lim x → 0 f ( x ) = f ( 0 ) = 1 \lim_{x \to 0} f(x) = f(0) = 1 lim x → 0 f ( x ) = f ( 0 ) = 1
答案 :函数在 x = 0 x = 0 x = 0 处连续。
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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2 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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