Continuity of a Function

A function $f$ is intuitively continuous at a point if its graph can be traced through that

point without lifting the pen -- there is no jump, no break, and no hole. Making this precise

needs exactly the machinery of limits: three separate conditions must all hold together.

Definition (continuity at a point). A function $f$ is said to be continuous at $x=a$ if

all three of the following hold:

1. $f(a)$ is defined (the function has a value at $a$);
2. $\displaystyle\lim_{x\to a}f(x)$ exists, i.e. the left-hand limit $\displaystyle\lim_{x\to a^-}f(x)$ and the right-hand limit $\displaystyle\lim_{x\to a^+}f(x)$ both exist and are equal;
3. $\displaystyle\lim_{x\to a}f(x) = f(a)$ -- the limit actually equals the function's value there.

Equivalently, $f$ is continuous at $a$ exactly when

If even one of these three conditions fails -- $f(a)$ is undefined, the limit does not exist

(left- and right-hand limits disagree), or the limit exists but does not match $f(a)$ -- then $f$

is said to be discontinuous at $a$, and $a$ is called a point of discontinuity.

Continuity on an interval. $f$ is continuous on an open interval $(p,q)$ if it is continuous

at every point of that interval; it is continuous on a closed interval $[p,q]$ if, in addition to

being continuous at every interior point, it is continuous from the right at $p$ and from the

left at $q$.

Algebra of continuous functions. If $f$ and $g$ are both continuous at $x=a$, then so are

$f+g$, $f-g$, $f\cdot g$, and $kf$ for any constant $k$; the quotient $f/g$ is continuous at $a$

provided $g(a)\neq0$. These follow directly from the algebra of limits, since continuity is

itself defined entirely in terms of a limit. A further, very useful fact is that the

composition of two continuous functions is continuous: if $g$ is continuous at $a$ and $f$ is

continuous at $g(a)$, then $f\circ g$ is continuous at $a$.

Which standard functions are continuous. Every polynomial function is continuous at every

real number (direct substitution always applies). A rational function $p(x)/q(x)$ is continuous

at every point where $q(x)\neq0$, and typically discontinuous where $q(x)=0$ -- unless the zero is

a removable one, cancelled by a matching factor in $p(x)$, in which case the limit still exists

even though $f(a)$ itself may be undefined or wrongly assigned. The functions $\sin x$ and

$\cos x$ are continuous for every real $x$; consequently $\tan x=\sin x/\cos x$ is continuous

except where $\cos x=0$, i.e. at odd multiples of $\pi/2$. The modulus function $f(x)=|x|$ is

continuous everywhere, including at $x=0$, even though (Section 2) it fails to be differentiable

there.

Worked illustration. For $f(x)=\begin{cases}x^2+1, & x\le2\\ 3x-1, & x>2\end{cases}$: here

$f(2)=2^2+1=5$; the left-hand limit is $\lim_{x\to2^-}(x^2+1)=5$; the right-hand limit is

$\lim_{x\to2^+}(3x-1)=3(2)-1=5$. Since all three agree, $f$ is continuous at $x=2$ -- exactly the

check Example 1 carries out in full.