Why a Function Can Have a Hole You Can Patch With One Point

A function is continuous when its graph can be drawn without lifting the pen — no jumps, gaps or holes. That simple picture has a precise definition using limits, and it matters because most results of calculus, from derivatives to integrals, work only for continuous functions.

This lesson covers continuity at a point and on an interval, the algebra of continuous functions, and types of discontinuity.

**A function f is continuous at $x=a$ when $f(a)$ is defined, ${\lim }_{x\to a}f(x)$ exists, and the two are equal; it is continuous on an interval when it is continuous at every point of that interval, using one-sided limits at closed end points.

Three conditions at $x=a$:**

- $f(a)$ exists
- The left-hand and right-hand limits are equal, so ${\lim }_{x\to a}f(x)$ exists
- ${\lim }_{x\to a}f(x)=f(a)$

Worked example. Let $f(x)=2x+1$ for $x\le 2$ and $f(x)=x^2+1$ for $x>2$. Is f continuous at 2?

- $f(2)=5$
- Left-hand limit $2(2)+1=5$; right-hand limit $4+1=5$
- All three agree, so f is continuous at 2

Worked example 2. Find k so that $f(x)=kx+3$ for $x<1$ and $f(x)=5x-1$ for $x\ge 1$ is continuous at 1. The left-hand limit is $k+3$ and $f(1)=4$, so $k+3=4$ and $k=1$.

On intervals. Polynomials, $\sin x$, $\cos x$ and $e^x$ are continuous everywhere; $\log x$ is continuous for $x>0$; $\tan x$ is continuous wherever $\cos x\ne 0$.

An everyday example. The temperature in Chennai through a day changes continuously — it cannot jump from 28 °C to 32 °C without passing through every value in between.

The substance. A function can be continuous on its whole domain and still have a break in its graph — $\frac{1}{x}$ is continuous at every point of its domain, because 0 is not in the domain.

**If f and g are continuous at a, then $f+g$, $f-g$, $fg$ and $kf$ are continuous at a, the quotient $\frac{f}{g}$ is continuous at a provided $g(a)\ne 0$, and a composite of continuous functions is continuous.

Why it works.** The limit laws carry over: $\lim (f+g)=\lim f+\lim g=f(a)+g(a)$.

Worked example. Where is $h(x)=\frac{x^2+\sin x}{x^2-5x+6}$ continuous?

- The numerator is a sum of continuous functions, so it is continuous everywhere
- The denominator $(x-2)(x-3)$ is continuous and is zero at $x=2$ and $x=3$
- So h is continuous on $\mathbb{R}-\{2,3\}$

Worked example 2. $e^{\cos x}$ is continuous everywhere, as a composite of $\cos x$ and $e^x$, and $|x-3|+x^3$ is continuous everywhere, as a sum of continuous functions.

An everyday example. The total weight of a water tanker being filled — the tank's own weight plus the water's weight — changes continuously because both parts do.

The substance. The converse fails — $f+g$ can be continuous when neither f nor g is, as with $[x]$ and $-[x]$, whose sum is 0 everywhere.

**A discontinuity at a is removable when ${\lim }_{x\to a}f(x)$ exists but $f(a)$ is undefined or different from the limit, so redefining one value makes f continuous; when the one-sided limits exist but differ it is a jump discontinuity, and when a one-sided limit fails to exist it is a discontinuity of the second kind.

Types of discontinuity:

- Removable** — the limit exists, but $f(a)$ is missing or wrong
- Jump, or first kind — left-hand and right-hand limits exist but are unequal
- Second kind — at least one one-sided limit does not exist, for example by growing without bound or oscillating

Worked example (removable). $f(x)=\frac{x^2-9}{x-3}$ is undefined at 3, but ${\lim }_{x\to 3}f(x)=6$. Defining $f(3)=6$ removes the discontinuity.

Worked example (jump). For $f(x)=[x]$ at $x=2$, the left-hand limit is 1 and the right-hand limit is 2, so the discontinuity cannot be removed.

Worked example (second kind). $\frac{1}{x-1}$ at $x=1$ has one-sided limits that grow without bound, and $\sin \frac{1}{x}$ at 0 oscillates without settling.

An everyday example. A parking fee that rises in steps at each completed hour has a jump discontinuity at every hour mark.

The substance. Only a removable discontinuity can be fixed by changing one value — at a jump, no single value of $f(a)$ can match two different one-sided limits.