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The Timpson Lyceum · Calculus 1 (2024)

Continuous Functions & Continuity at a Point - Lecture 6 (Calculus 1)

Overview

This lecture introduces continuity, the second of calculus's four big ideas (limits, continuity, derivatives, integrals). Building directly on the precise definition of a limit, it defines what it means for a function to be continuous at a point, then extends that idea to endpoints of a closed interval and to whole functions. It catalogs the common types of discontinuity (removable, jump, infinite, oscillating), shows that the usual arithmetic and composition rules for limits carry over to continuity, and finishes with two important applications: the Intermediate Value Theorem and continuous extensions. Because nearly every function we work with is continuous except at a handful of strange points, the goal is to recognize and reason about those exceptional points.

Key Concepts

Continuity at a point (simple form): A function ff is continuous at x=cx = c if and only if lim⁡x→cf(x)=f(c).\lim_{x \to c} f(x) = f(c). Three things must all hold: f(c)f(c) is defined, lim⁡x→cf(x)\lim_{x \to c} f(x) exists, and the two are equal. This is just the limit definition with the target value LL replaced by the function's actual value f(c)f(c).

Epsilon–delta form: ff is continuous at cc if for every ε>0\varepsilon > 0 there exists δ>0\delta > 0 such that whenever ∣x−c∣<δ|x - c| < \delta, then ∣f(x)−f(c)∣<ε|f(x) - f(c)| < \varepsilon. Identical to the limit definition except LL becomes f(c)f(c), so the function approaches its own value at the point.

Why a limit can fail to equal the value:

Continuity at endpoints: For a function defined on a closed interval [a,b][a, b], one-sided limits are used at the ends:

Continuous on its domain: A function is continuous if it is continuous at every point in its domain. For example, f(x)=1/xf(x) = 1/x is a continuous function even though it blows up at 00 — because 00 is not in its domain.

Types of discontinuity:

Continuity laws (inherited from limit laws): If ff and gg are continuous at cc, then so are:

Consequences: Every polynomial is continuous everywhere (lim⁡x→cP(x)=P(c)\lim_{x \to c} P(x) = P(c)); every rational function P(x)/Q(x)P(x)/Q(x) is continuous wherever Q(c)≠0Q(c) \neq 0; and the absolute value function is continuous on all of R\mathbb{R}. Combined with composition, this shows expressions like x2−2x+5\sqrt{x^2 - 2x + 5} or ∣x−2∣/(x2−2)|x - 2|/(x^2 - 2) are continuous on their natural domains.

Limit-inside-composition theorem: If gg is continuous at bb and lim⁡x→cf(x)=b\lim_{x \to c} f(x) = b, then lim⁡x→cg(f(x))=g(lim⁡x→cf(x))=g(b).\lim_{x \to c} g\big(f(x)\big) = g\Big(\lim_{x \to c} f(x)\Big) = g(b). The corollary that composition preserves continuity follows immediately when b=f(c)b = f(c).

Intermediate Value Theorem (IVT): If ff is continuous on a closed interval [a,b][a, b] and y0y_0 is any value between f(a)f(a) and f(b)f(b), then there exists at least one cc in [a,b][a, b] with f(c)=y0f(c) = y_0. It guarantees a solution exists (at least one cc) without locating it; commonly used to confirm an equation has a root before approximating it.

Continuous extension: Given a function with a removable hole, define a new function that agrees everywhere except at the bad point, where it takes the value of the limit. This "fills the hole," producing a function that is continuous there. In practice you either set the value equal to the limit at that point, or algebraically cancel the common factor causing the hole.

Worked Example

Use the IVT to show that x3−x−1=0x^3 - x - 1 = 0 has a solution. Let f(x)=x3−x−1f(x) = x^3 - x - 1, which is a polynomial and therefore continuous on every closed interval. Evaluate the endpoints, then invoke the IVT with the target value y0=0y_0 = 0:

f(1)=13−1−1=−1,f(2)=23−2−1=8−2−1=5,f(1)=−1  <  0  <  5=f(2)  ⇒  ∃ c∈[1,2] with f(c)=0.\begin{aligned} f(1) &= 1^3 - 1 - 1 = -1, \\ f(2) &= 2^3 - 2 - 1 = 8 - 2 - 1 = 5, \\ f(1) = -1 \;&<\; 0 \;<\; 5 = f(2) \;\Rightarrow\; \exists\, c \in [1, 2] \text{ with } f(c) = 0. \end{aligned}

Since ff changes sign from −1-1 at x=1x = 1 to 55 at x=2x = 2 and is continuous, the IVT guarantees some cc in [1,2][1, 2] where f(c)=0f(c) = 0 — a root of the equation. (A numeric method could then narrow in on its value.)

As a second, shorter illustration, a continuous extension of f(x)=x2+x−6x2−4f(x) = \dfrac{x^2 + x - 6}{x^2 - 4} at x=2x = 2 comes from factoring and canceling the offending factor:

x2+x−6x2−4=(x+3)(x−2)(x−2)(x+2)=x+3x+2,F(2)=2+32+2=54,\begin{aligned} \frac{x^2 + x - 6}{x^2 - 4} &= \frac{(x + 3)(x - 2)}{(x - 2)(x + 2)} = \frac{x + 3}{x + 2}, \\ F(2) &= \frac{2 + 3}{2 + 2} = \frac{5}{4}, \end{aligned}

so defining F(x)=x+3x+2F(x) = \dfrac{x + 3}{x + 2} fills the hole, and FF is continuous at x=2x = 2.

Summary