Continuity — Drawing Without Lifting the Pen

Informally, a function is continuous at a point if its graph can be drawn through that point without lifting the pen. Two piecewise examples show how this can fail: a function that jumps from value $1$ to value $2$ across $x = 0$, and a function that equals $1$ everywhere except for a lone value $2$ at $x = 0$. In both cases the pen must lift.

Definition. Let $c$ be a point in the domain of $f$. Then $f$ is continuous at $c$ if

$$\boxed{\;\lim_{x \to c} f(x) = f(c)\;}$$

Unpacked, three things must hold and agree: the left hand limit, the right hand limit, and the value $f(c)$. If any is missing or they disagree, $f$ is discontinuous at $c$, and $c$ is a point of discontinuity.

A function is continuous (with no point named) if it is continuous at every point of its domain. Subtle but exam-relevant: $f(x) = \frac{1}{x}$ is a continuous function — $0$ is not in its domain, so there is no domain point where continuity fails.

The case-analysis routine for piecewise functions

For $f(x) = \begin{cases} x + 2, & x \leq 1 \\ x - 2, & x > 1 \end{cases}$ check three cases: away from the joint on each side, each piece is a polynomial (continuous); at the joint $x = 1$: LHL $= 1 + 2 = 3$, RHL $= 1 - 2 = -1$. They disagree, so $x = 1$ is the only point of discontinuity. Every piecewise problem in Exercise territory is this routine — the only work is at the joints.

Algebra of continuous functions

If $f$ and $g$ are continuous at $c$, then so are $f + g$, $f - g$, $f \cdot g$, and $\frac{f}{g}$ (wherever $g(c) \neq 0$). Combined with two seed facts — constants and the identity $f(x) = x$ are continuous — this yields whole families at once:

1. Every polynomial is continuous (sums of products of $x$ with constants).
2. Every rational function $\frac{p(x)}{q(x)}$ is continuous on its domain (quotient rule of the algebra).
3. $\sin x$ and $\cos x$ are continuous everywhere; hence $\tan x = \frac{\sin x}{\cos x}$ is continuous except where $\cos x = 0$, i.e. at odd multiples of $\frac{\pi}{2}$.
4. $\vert x \vert$ is continuous everywhere — including the corner at $0$ (both one-sided limits are $0 = f(0)$; a corner bends the graph but never breaks it).
5. Composites: if $f$ and $g$ are continuous, so is $f \circ g$ — so $\sin(x^2)$, $\vert 1 - x + \vert x \vert \vert$ and friends are continuous with no computation.

One famous troublemaker: the greatest integer function $[x]$ is discontinuous at every integer — at each integer the left hand limit sits one step below the value.

Solved Examples

Example 1 — A first check from the definition

Check the continuity of $f(x) = 2x + 3$ at $x = 1$.

Step 1 — the limit: $\lim_{x \to 1} (2x + 3) = 2(1) + 3 = 5$.

Step 2 — the value: $f(1) = 5$.

Answer: limit $=$ value, so $f$ is continuous at $x = 1$.

Example 2 — A two-piece function with a jump

Find all points of discontinuity of $f(x) = \begin{cases} 2x + 3, & x \leq 2 \\ 2x - 3, & x > 2 \end{cases}$.

Step 1 — away from the joint: each piece is a polynomial, continuous on its open piece.

Step 2 — at $x = 2$: LHL $= 2(2) + 3 = 7$; RHL $= 2(2) - 3 = 1$.

Answer: the one-sided limits disagree, so $x = 2$ is the only point of discontinuity.

Example 3 — Three pieces, two joints

Find all points of discontinuity of $f(x) = \begin{cases} \vert x \vert + 3, & x \leq -3 \\ -2x, & -3 < x < 3 \\ 6x + 2, & x \geq 3 \end{cases}$.

Step 1 — joint $x = -3$: LHL $= \vert -3 \vert + 3 = 6$; RHL $= -2(-3) = 6$; value $f(-3) = 6$. All agree — continuous here.

Step 2 — joint $x = 3$: LHL $= -2(3) = -6$; RHL $= 6(3) + 2 = 20$. Disagree.

Answer: discontinuous only at $x = 3$. Always check every joint — one may pass while another fails.

Example 4 — A lone bad value

Discuss the continuity of $f(x) = \begin{cases} x + 2, & x < 1 \\ 0, & x = 1 \\ x - 2, & x > 1 \end{cases}$ at $x = 1$.

Step 1 — one-sided limits: LHL $= 3$, RHL $= -1$ — already unequal, so no value of $f(1)$ could rescue continuity.

Answer: discontinuous at $x = 1$ (and continuous everywhere else). Redefining a single value can only fix a discontinuity when the two one-sided limits already agree.

Example 5 — Why every polynomial is continuous

Show that every polynomial function is continuous.

Step 1 — seed facts: constant functions and the identity $f(x) = x$ are continuous (each limit is immediate from the definition).

Step 2 — build up with the algebra: $x^k$ is a repeated product of continuous functions, $a_k x^k$ a product with a constant, and a polynomial $a_0 + a_1 x + \cdots + a_n x^n$ a finite sum of these.

Answer: by the algebra of continuous functions, every polynomial is continuous at every real number. ∎ The identical argument makes every rational function continuous on its domain.

Example 6 — The greatest integer function

Find all points of discontinuity of $f(x) = [x]$, the greatest integer less than or equal to $x$.

Step 1 — away from integers: on any interval between consecutive integers, $[x]$ is constant — continuous.

Step 2 — at an integer $n$: LHL $= n - 1$ (values just below $n$ round down to $n-1$), while RHL $= n = f(n)$.

Answer: $[x]$ is discontinuous at every integer and continuous everywhere else — an infinite family of jump points from one innocent-looking definition.