Continuity: The Unbroken Path
Intuitively, a function is continuous if you can draw its graph without lifting your pen from the paper. Mathematically, this means that as \( x \) gets closer to a point \( x_0 \), the function value \( f(x) \) must get closer to \( f(x_0) \).
The formal definition uses the same \(\epsilon-\delta\) language we saw with limits, because continuity at \(x_0\) is essentially saying \(\lim_{x \to x_0} f(x) = f(x_0)\).
Interactive \(\epsilon-\delta\) Definition
For a function to be continuous at \( x_0 \): For any "target error" \(\epsilon \gt 0\) (height of the blue box), we must be able to find a "tolerance" \(\delta \gt 0\) (width of the green box) such that the entire graph within the green width stays inside the blue height.
Function Type
Formal Definitions
Metric Space Definition
Let \((X, d_X)\) and \((Y, d_Y)\) be metric spaces. A function \(f: X \to Y\) is continuous at \(x_0 \in X\) if:
\(d_X(x, x_0) \lt \delta \implies d_Y(f(x), f(x_0)) \lt \epsilon\)
Sequence Definition
A function \(f: X \to Y\) is continuous at \(x_0\) if and only if for every sequence \((x_n)\) in \(X\) converging to \(x_0\):
This is often easier to use for disproving continuity (find one sequence that breaks it!).
Practice Problems
True/False
1. If \(f\) is continuous at \(x_0\), then \(\lim_{x \to x_0} f(x)\) exists.
True. Continuity implies the limit exists and equals the function value.
2. A function can be continuous at a point where it is not defined.
False. For \(f\) to be continuous at \(x_0\), \(f(x_0)\) must be defined.
3. Polynomial functions are continuous everywhere.
True. Polynomials are continuous for all real numbers.
Fill in the Blank
1. A function \(f\) is continuous at \(x_0\) if \(\lim_{x \to x_0} f(x) = \) ______.
\(f(x_0)\).
2. If \(\lim_{x \to x_0^-} f(x) \neq \lim_{x \to x_0^+} f(x)\), the function has a ______ discontinuity at \(x_0\).
Jump.
3. The ______ Value Theorem states that if \(f\) is continuous on \([a,b]\) and \(y\) is between \(f(a)\) and \(f(b)\), there exists \(c \in (a,b)\) such that \(f(c) = y\).
Intermediate.
Full Problems
1. Determine if \(f(x) = \begin{cases} \frac{x^2-1}{x-1} & x \neq 1 \\ 2 & x = 1 \end{cases}\) is continuous at \(x=1\).
Yes. \(\lim_{x \to 1} \frac{x^2-1}{x-1} = \lim_{x \to 1} (x+1) = 2\). Since the limit equals \(f(1)\), it is continuous.
2. Find the value of \(k\) that makes \(f(x) = \begin{cases} kx^2 & x \le 2 \\ 2x + k & x \gt 2 \end{cases}\) continuous at \(x=2\).
We need \(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)\).
\(k(2)^2 = 2(2) + k \implies 4k = 4 + k \implies 3k = 4 \implies k = 4/3\).
3. Prove that the equation \(x^3 - x - 1 = 0\) has a root in the interval \([1, 2]\).
Let \(f(x) = x^3 - x - 1\). \(f\) is continuous. \(f(1) = 1 - 1 - 1 = -1 \lt 0\). \(f(2) = 8 - 2 - 1 = 5 \gt 0\). By the Intermediate Value Theorem, there exists \(c \in (1,2)\) such that \(f(c) = 0\).
4. Classify the discontinuity of \(f(x) = \frac{|x|}{x}\) at \(x=0\).
Jump Discontinuity. \(\lim_{x \to 0^+} \frac{x}{x} = 1\) and \(\lim_{x \to 0^-} \frac{-x}{x} = -1\). The limits exist but are not equal.