1.11 Defining Continuity at a Point
Keywords
Continuity 连续性 • Existence 存在性
Point Discontinuity 点间断 • Three-Step Test 连续性三步检验
The Mathematical Definition
A function \(f(x)\) is continuous at a point \(x=c\) if the graph is "unbroken" there. However, to prove this analytically, we must satisfy three specific conditions.
核心要点：The Three Conditions of Continuity
A function \(f(x)\) is continuous at \(x=c\) if and only if:
-
\(f(c)\) exists. (The function is defined at \(c\); there is no hole or asymptote at the point).
-
\(\lim\limits_{x \to c} f(x)\) exists. (The left-hand limit equals the right-hand limit).
-
\(\lim\limits_{x \to c} f(x) = f(c)\). (The limit matches the actual function value).
If any of these three conditions are not met, the function is discontinuous at \(x=c\).
Visualizing the Conditions
Let's look at a function that is continuous at \(x = 2\).
- Condition 1: \(f(2) = 2\) (The solid point exists).
- Condition 2: \(\lim\limits_{x \to 2} f(x) = 2\) (The path from both sides leads to \(y=2\)).
- Condition 3: The limit (\(2\)) equals the point (\(2\)).
- Result: Continuous at \(x=2\).
Piecewise Continuity: Solving for a Constant
A classic AP-style problem asks you to find a value for a constant (like \(k\)) that makes a piecewise function continuous.
Example: Find the value of \(k\) that makes \(f(x)\) continuous at \(x=3\): $$ f(x) = \begin{cases} x^2 - 1, & x \le 3 \ kx + 2, & x > 3 \end{cases} $$
To be continuous, the limit from the left must equal the limit from the right:
-
Left-hand limit: \(\lim\limits_{x \to 3^-} (x^2 - 1) = 3^2 - 1 = 8\).
-
Right-hand limit: \(\lim\limits_{x \to 3^+} (kx + 2) = 3k + 2\).
-
Set them equal: \(8 = 3k + 2 \implies 6 = 3k \implies k = 2\).
Extension: Continuity on an Interval
We say a function is continuous on an open interval \((a, b)\) if it is continuous at every point in that interval. For a closed interval \([a, b]\), we also check the endpoints using one-sided limits: \(\lim\limits_{x \to a^+} f(x) = f(a)\) and \(\lim\limits_{x \to b^-} f(x) = f(b)\).