Continuity

A function $f$ is continuous at $a$ when

1. $f(a)$ is defined;
2. $\lim_{x\to a}f(x)$ exists;
3. $\lim_{x\to a}f(x)=f(a)$.

Thus continuity joins two kinds of information: the function's nearby behavior and its assigned value at the point.

One-Sided and Interval Continuity

At an endpoint, only the side contained in the domain is relevant. A function is continuous on $[a,b]$ when it is continuous on $(a,b)$, right-continuous at $a$, and left-continuous at $b$:

$$\lim_{x\to a^+}f(x)=f(a), \qquad \lim_{x\to b^-}f(x)=f(b).$$

Typical discontinuities include a removable hole, a jump between unequal one-sided limits, and unbounded behavior near a vertical asymptote.

Open full-size image

Compare the open circle, the filled point, and the values approached from each side. In (a), changing the single assigned value can restore continuity. In (b), the two one-sided limits disagree; in (c), the values become unbounded. Neither can be repaired by changing just f(a). Access for free at OpenStax.

Gilbert Strang & Edwin Herman, OpenStax, Calculus Volume 1, Figure 2.37. · CC BY-NC-SA 4.0

Functions Built from Continuous Functions

Polynomials are continuous on $\mathbb{R}$. Rational functions are continuous where their denominators are nonzero. Exponential, logarithmic, trigonometric, and root functions are continuous on their respective domains.

Sums, products, and valid quotients of continuous functions remain continuous. If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then

$$(f\circ g)(x)=f(g(x))$$

is continuous at $a$. This justifies moving the limit through $f$ under those continuity hypotheses:

$$\lim_{x\to a}f(g(x)) =f\left(\lim_{x\to a}g(x)\right).$$

Intermediate Value Theorem

If $f$ is continuous on $[a,b]$ and $N$ lies between $f(a)$ and $f(b)$, then there is at least one $c\in[a,b]$ such that $f(c)=N$.

The theorem guarantees existence, not uniqueness or a formula for $c$. In particular, if $f(a)$ and $f(b)$ have opposite signs, at least one root lies in $[a,b]$.

Repairing a hole and locating a root

Let $f(x)=(x^2-1)/(x-1)$ for $x\ne1$ and $f(1)=c$. Near $1$, cancellation gives $f(x)=x+1$, so the limit is $2$. Exactly the choice $c=2$ makes $f$ continuous at $1$; changing one value can repair a hole but cannot repair a jump.

For $p(x)=x^3-x-1$, continuity and $p(1)=-1$, $p(2)=5$ guarantee a root in $(1,2)$. Bisection retains the half-interval whose endpoint values have opposite signs; since $p(1.5)=0.875$, the first retained interval is $[1,1.5]$. After $n$ halvings its width is $2^{-n}$, so the midpoint is within $2^{-n-1}$ of a root. Continuity is essential: $1/x$ has opposite signs at $-1$ and $1$ but no zero, because it is not continuous on $[-1,1]$.

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