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Calculus BC · Unit 1 · 1.12
AP Calculus · Limits and Continuity
Confirming Continuity over an Interval
Every interior point, plus one side at each endpoint — and every seam checked separately.
key idea$\text{Continuous on }[a,b]:\ \text{every interior point, plus ONE side at each end}$
What it says
Continuity on an interval
A function is continuous on an open interval $(a,b)$ if it is continuous at every point inside. On a closed interval $[a,b]$ you additionally need one-sided continuity at the endpoints: $\lim_{x\to a^{+}}f=f(a)$ and $\lim_{x\to b^{-}}f=f(b)$.
Why endpoints only get one side
At $x=a$ there is nothing to the left of the interval, so demanding a two-sided limit would be unreasonable. $\sqrt{x}$ is continuous on $[0,\infty)$ even though $\sqrt{x}$ is undefined for $x\lt 0$ — the left side is simply not part of the question.
| Family | Continuous on | Where it breaks |
|---|
| polynomials | all reals | nowhere |
| rational $\dfrac{p}{q}$ | all reals except zeros of $q$ | denominator $=0$ |
| $\sqrt[\,\text{even}\,]{\;}$ | where the radicand $\ge 0$ | radicand negative |
| $\ln x$ | $(0,\infty)$ | $x\le 0$ |
| $e^{x}$, $\sin x$, $\cos x$ | all reals | nowhere |
| $\tan x$ | between consecutive odd multiples of $\tfrac{\pi}{2}$ | $x=\tfrac{\pi}{2}+k\pi$ |
| piecewise | each piece's own domain | possibly at the seams — must be checked |
Quoting the family is a legitimate justification: “$f$ is a polynomial, therefore continuous on $[a,b]$” is a complete argument.
Reading the notation
$(a,b)$ openEndpoints excluded. Only interior points are tested.
$[a,b]$ closedEndpoints included, each tested with a one-sided limit.
$\lim_{x\to a^{+}}f=f(a)$Right-continuity at the left endpoint — the only condition asked of $a$.
“Continuous on its domain”A weaker, often safer phrasing. $\tfrac1x$ is continuous on its domain, but it is not continuous on $[-1,1]$, because $0$ is in that interval and not in the domain.
Union of intervalsWhere a function breaks at isolated points, report the answer as a union, e.g. $(-\infty,2)\cup(2,\infty)$ — never as a single interval containing the break.
Worked examples
EX 1
Name the intervals
$\dfrac{x+1}{x^{2}-4}$On which intervals is $f(x)=\dfrac{x+1}{x^{2}-4}$ continuous?
1
Identify the family
A rational function, so it is continuous everywhere its denominator is non-zero.
2
Find the breaks
$x^{2}-4=(x-2)(x+2)=0$ at $x=2$ and $x=-2$.
3
Check for cancellation
The numerator $x+1$ shares neither factor, so both breaks are genuine.
4
Write the answer as a union
$(-\infty,-2)\cup(-2,2)\cup(2,\infty)$.
5
Common slip
Writing “continuous for all $x\neq\pm2$” is acceptable, but if the question says intervals, give the union.
EX 2
Endpoints get one side
$\sqrt{9-x^{2}}$Show that $g(x)=\sqrt{9-x^{2}}$ is continuous on $[-3,3]$.
1
Find the domain
The radicand needs $9-x^{2}\ge0$, i.e. $-3\le x\le3$. The domain is exactly $[-3,3]$.
2
Interior points
On $(-3,3)$ the radicand is strictly positive, and a root of a continuous positive function is continuous. So $g$ is continuous at every interior point.
3
Left endpoint
$\lim_{x\to-3^{+}}\sqrt{9-x^{2}}=0=g(-3)$. Right-continuous at $-3$.
4
Right endpoint
$\lim_{x\to3^{-}}\sqrt{9-x^{2}}=0=g(3)$. Left-continuous at $3$.
5
Conclude
All interior points plus both one-sided endpoint conditions hold, so $g$ is continuous on the closed interval $[-3,3]$.
EX 3
Check every seam
piecewiseIs $h(x)=\begin{cases}x^{2},& x\lt 1\\ 2x-1,& 1\le x\lt 4\\ x+3,& x\ge 4\end{cases}$ continuous on $[0,6]$?
1
Each piece alone
All three pieces are polynomials, so the only possible failures are at the seams $x=1$ and $x=4$.
2
Seam at $x=1$
Left: $1^{2}=1$. Right: $2(1)-1=1$. Value: $h(1)=1$. All agree, so $h$ is continuous at $1$.
3
Seam at $x=4$
Left: $2(4)-1=7$. Right: $4+3=7$. Value: $h(4)=7$. All agree.
4
Endpoints of $[0,6]$
$0$ and $6$ sit inside the first and third pieces respectively, where each is a polynomial — so the one-sided conditions hold trivially.
5
Conclude
$h$ is continuous on all of $[0,6]$. Every seam had to be checked; passing one says nothing about the other.
EX 4
A break inside the interval
$\tan x$Is $\tan x$ continuous on $[0,\pi]$?
1
Recall where $\tan$ fails
$\tan x=\dfrac{\sin x}{\cos x}$ is undefined wherever $\cos x=0$, i.e. at $x=\tfrac{\pi}{2}+k\pi$.
2
Test the interval
$\tfrac{\pi}{2}$ lies inside $[0,\pi]$.
3
Conclude
No. $\tan x$ is not continuous on $[0,\pi]$ — it is not even defined at $\tfrac{\pi}{2}$.
4
Give the correct intervals
$\left[0,\tfrac{\pi}{2}\right)\cup\left(\tfrac{\pi}{2},\pi\right]$.
5
Why this matters
Theorems that require continuity on a closed interval simply do not apply here, no matter how well-behaved the function looks elsewhere.
Boundary cases
Endpoints need only one side
Demanding a two-sided limit at $a$ or $b$ would make almost every closed-interval claim fail. $\sqrt{x}$ on $[0,4]$ is the standard example.
“Continuous on its domain” ≠ “continuous on $[a,b]$”
$\tfrac1x$ is continuous on its domain but not on $[-1,1]$, because $0$ lies in the interval and outside the domain.
Report unions, not single intervals
$(-\infty,2)\cup(2,\infty)$ is right; $(-\infty,\infty)$ excluding a point is not an interval.
Every seam, not just the first
A piecewise function with three pieces has two seams and both must be checked independently.