Continuity at a Point and on an Interval
Define continuity using three conditions and classify discontinuities as removable, jump, or infinite. Apply the Intermediate Value Theorem to guarantee the existence of zeros and apply to real-world existence proofs.
A lesson of Calculus, in Limits and Continuity, about 10 minutes.
By the end of this lesson, you will be able to apply the three-part definition of continuity, classify discontinuities, determine the continuity of composite functions, and apply the Intermediate Value and Extreme Value Theorems.
Three Conditions for Continuity at $c$
- $f(c)$ is defined
- $\lim_{x \to c}f(x)$ exists
- $\lim_{x \to c}f(x) = f(c)$
All three must hold. The limit and function value must match.
Types of Discontinuity
Removable: $\lim_{x\to c}$ exists, but either $f(c)$ is undefined or $f(c) \neq L$. The graph has a hole. Can be fixed by redefining $f(c) = L$.
Jump: one-sided limits exist but are unequal ($L^- \neq L^+$). The graph jumps.
Infinite: $f(x) \to \pm\infty$ as $x \to c$. Vertical asymptote. Cannot be fixed.
Oscillatory: $f$ oscillates near $c$ (like $\sin(1/x)$ at 0). Cannot be fixed.
Continuity of Standard Functions
Continuous everywhere on their domains:
- Polynomials
- Rational functions (where denominator ≠ 0)
- Trig functions ($\sin, \cos$ everywhere; $\tan$ where $\cos \neq 0$)
- $e^x$, $\ln x$ (on $(0,\infty)$)
- $\sqrt[n]{x}$ (even $n$: on $[0,\infty)$; odd $n$: everywhere)
Continuity is preserved under: addition, subtraction, multiplication, division (denominator ≠ 0), composition.
Continuity of Compositions
Theorem: if $g$ is continuous at $c$ and $f$ is continuous at $g(c)$, then $f \circ g$ is continuous at $c$.
$$\lim_{x \to c}f(g(x)) = f\left(\lim_{x \to c}g(x)\right) = f(g(c))$$
This lets us move limits "through" continuous functions.
The Intermediate Value Theorem (IVT)
If $f$ is continuous on $[a,b]$ and $N$ is any value between $f(a)$ and $f(b)$, then there exists at least one $c \in (a,b)$ such that $f(c) = N$.
Corollary (Existence of Zeros): If $f$ is continuous on $[a,b]$ and $f(a)$ and $f(b)$ have opposite signs, then $f$ has at least one zero in $(a,b)$.
The Extreme Value Theorem (EVT)
If $f$ is continuous on $[a,b]$, then $f$ attains both an absolute maximum and an absolute minimum on $[a,b]$.
The EVT requires: (1) continuity, (2) closed and bounded interval.
Example 1 — Applying IVT
Show that $f(x) = x^5 - 3x + 1$ has a zero in $[1, 2]$.
$f(1) = 1 - 3 + 1 = -1 < 0$