Algebra of continuous functions

Just as limits combine algebraically (Theorem 9.2), so does continuity — which should be unsurprising, since continuity is itself defined directly in terms of limits. If $f$ and $g$ are both continuous at $x_0$, then so are:

1. $f+g$,
2. $f-g$,
3. $f\cdot g$, and
4. $f/g$, provided $g(x_0)\ne0$.

A fifth, independent rule handles composition:

5. Composite function rule. If $g$ is continuous at $x_0$ and $f$ is continuous at $g(x_0)$, then the composite $f\circ g$ is continuous at $x_0$.

Continuity on a closed interval. Continuity "at a point" needs an open interval surrounding that point (so both sides can be approached) — but a closed interval $[a,b]$ has no room to spare outside its endpoints $a$ and $b$. This is handled with a one-sided adjustment:

Definition 9.9. $f:[a,b]\to\mathbb R$ is continuous on $[a,b]$ if it is continuous on the open interval $(a,b)$, and $\lim_{x\to a^+}f(x)=f(a)$, and $\lim_{x\to b^-}f(x)=f(b)$ — i.e. continuous from the right at $a$ and from the left at $b$.

Illustration 9.7. $f(x)=\sqrt{1-x^2}$ has domain exactly $[-1,1]$ (since $1-x^2\ge0$ only there). At any interior point $c\in(-1,1)$, $\lim_{x\to c} f(x)=\sqrt{1-c^2}=f(c)$ by the continuity of the square-root and polynomial building blocks (or, equivalently, by the composite rule, since $1-x^2$ is a continuous polynomial and $\sqrt{\phantom x}$ is continuous on its domain). At the two endpoints, the one-sided limits also match the function value: $\lim_{x\to-1^+}f(x)=0=f(-1)$ and $\lim_{x\to1^-}f(x)=0=f(1)$. So $f$ is continuous on the entire closed interval $[-1,1]$.

Worked techniques (Example 9.37).

• $f(x)=\tan x$ is continuous on each open interval between consecutive points where it is undefined, i.e. on each interval of the form $\left(\left(n-\tfrac12\right)\pi,\left(n+\tfrac12\right)\pi\right)$ for integer $n$ — continuity is a local property, so a function can be perfectly continuous "in pieces" even though it is undefined at isolated points scattered through its natural domain.
• $g(x)=\sin(1/x)$ for $x\ne0$, with $g(0)=0$: away from $0$ this is continuous (a composite of continuous functions), but at $0$ it fails, because $\lim_{x\to0}\sin(1/x)$ does not exist at all — the function oscillates through every value in $[-1,1]$ infinitely often as $x\to0$ — so no choice of value at $0$ could ever make it continuous there.
• $h(x)=x\sin(1/x)$ for $x\ne0$, with $h(0)=0$: this one is continuous even at $0$, because although $\sin(1/x)$ itself has no limit, it is always bounded between $-1$ and $1$, so $-|x|\le x\sin(1/x)\le|x|$; both bounds tend to $0$ as $x\to0$, and the Sandwich Theorem forces $\lim_{x\to0}h(x)=0=h(0)$. This example nicely shows that multiplying an oscillating-but-bounded factor by a factor shrinking to $0$ can restore continuity where the oscillating factor alone had none. …