Families of Continuous Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 2.4: “Continuity” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Key idea
Direct substitution works because the functions you encounter in calculus are almost always built from a small collection of continuous building blocks. Knowing which families are continuous -- and on exactly which domain -- means you can evaluate a wide range of limits by inspection rather than by building a table or going back to the definition.
The key list is short: polynomials, rational functions (on their domains), root functions (where defined), trigonometric functions (on their domains), exponential functions, and logarithms. Anything built from these by addition, multiplication, division, and composition is continuous wherever it is defined.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
| Function family | Continuous on... | Notes |
|---|---|---|
| Polynomials | $(-\infty, \infty)$ | Always |
| Rational $p(x)/q(x)$ | $\{x : q(x) \neq 0\}$ | Discontinuous where denominator is zero |
| $\sqrt[n]{x}$ (odd $n$) | $(-\infty, \infty)$ | Cube root, fifth root, etc. |
| $\sqrt{x}$ (even root) | $[0, \infty)$ | Or wherever argument is non-negative |
| $\sin x$, $\cos x$ | $(-\infty, \infty)$ | Always |
| $\tan x$ | $\{x : \cos x \neq 0\}$ | Excludes $x = \pi/2 + n\pi$ |
| $\csc x$, $\sec x$, $\cot x$ | Excludes zeros of $\sin$ or $\cos$ | As appropriate |
| $e^x$ | $(-\infty, \infty)$ | Always |
| $a^x$ ($a > 0$, $a \neq 1$) | $(-\infty, \infty)$ | Always |
| $\ln x$ | $(0, \infty)$ | Positive reals only |
| $\log_a x$ ($a > 0$, $a \neq 1$) | $(0, \infty)$ | Positive reals only |
Key Concepts
1. Polynomials: Continuous Everywhere
A polynomial $p(x) = a_n x^n + \cdots + a_1 x + a_0$ is continuous on $(-\infty, \infty)$.
Why. The function $f(x) = c$ is continuous everywhere (constant limit). The function $f(x) = x$ is continuous everywhere ($\lim_{x \to a} x = a$). By the product law, $x^2 = x \cdot x$ is continuous; by induction, $x^n$ is continuous. Multiplying by a constant (constant multiple law) and summing finitely many terms (sum law) preserves continuity. So every polynomial is continuous everywhere.
Consequence. $\lim_{x \to a} p(x) = p(a)$ for every polynomial $p$ and every real number $a$.
2. Rational Functions: Continuous on Their Domain
A rational function $r(x) = p(x)/q(x)$ is continuous at every $a$ where $q(a) \neq 0$.
Why. By the quotient law, a ratio of two continuous functions is continuous wherever the denominator is nonzero.
What happens at zeros of $q$. Either:
- $p(a) \neq 0$: the denominator drives to zero while the numerator does not; vertical asymptote.
- $p(a) = 0$ and $q(a) = 0$: indeterminate form $\frac{0}{0}$; factor and cancel to determine whether there is a hole or an asymptote.
Example 1. $r(x) = \dfrac{x^2 + 1}{x - 3}$ is continuous for all $x \neq 3$.
Example 2. $r(x) = \dfrac{x^2 - 9}{x - 3} = x + 3$ (for $x \neq 3$) has a removable discontinuity at $x = 3$: the function is undefined there, but the limit is 6.
3. Root Functions
The square root $\sqrt{x}$ is continuous on $[0, \infty)$: defined and continuous for $x > 0$, and right-continuous at $x = 0$.
More generally, $\sqrt[n]{g(x)}$ is continuous wherever $g(x) \geq 0$ (for even $n$) or wherever $g(x)$ is defined (for odd $n$).
Example 3. $f(x) = \sqrt{x^2 - 4}$ is continuous on $(-\infty, -2] \cup [2, \infty)$ (where $x^2 - 4 \geq 0$).
4. Trigonometric Functions
$\sin x$ and $\cos x$ are continuous on $(-\infty, \infty)$. This follows from a geometric argument using the squeeze theorem.
Derived trig functions:
- $\tan x = \sin x / \cos x$: continuous wherever $\cos x \neq 0$, i.e., $x \neq \pi/2 + n\pi$.
- $\cot x = \cos x / \sin x$: continuous wherever $\sin x \neq 0$, i.e., $x \neq n\pi$.
- $\sec x = 1/\cos x$: same domain as $\tan x$.
- $\csc x = 1/\sin x$: same domain as $\cot x$.
Example 4. $\lim_{x \to \pi/3} \tan x = \tan(\pi/3) = \sqrt{3}$. (Direct substitution, since $\pi/3 \neq \pi/2 + n\pi$.)
5. Exponential and Logarithmic Functions
$e^x$ and, more generally, $a^x$ for any positive base $a \neq 1$ are continuous on $(-\infty, \infty)$.
$\ln x$ is continuous on $(0, \infty)$. Similarly, $\log_a x$ is continuous on $(0, \infty)$.
Example 5. $\lim_{x \to 2} e^{x^2 - 3} = e^{4-3} = e^1 = e$.
Example 6. $\lim_{x \to 1} \ln(x^2 + 1) = \ln(2)$.
6. Where this shows up
Each function family above is continuous on its natural domain. Composites, sums, products, and quotients of these families are continuous wherever defined (by the continuity laws). So for any standard function $f$ encountered in calculus:
$\lim_{x \to a} f(x) = f(a)$ whenever $f$ is defined at $a$ and built from the families above.
This is the exact reason direct substitution is the first technique to try for any limit.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Claiming $\ln x$ is continuous everywhere | “$\lim_{x \to 0} \ln x$ can be found by plugging in” | $\ln$ is only continuous for $x > 0$; $\ln 0$ is undefined and the limit is $-\infty$ |
| Forgetting $\tan$ has excluded points | “Plugging in $\pi/2$” into $\tan x$ | $\tan(\pi/2)$ is undefined; $\pi/2$ is excluded from the domain |
| Using continuity where numerator and denominator both vanish | “$1/(x-1)$ is continuous at $x=1$” | The denominator is zero at $x=1$; the function is undefined and has a vertical asymptote |
Leveled Practice
Level 1 -- Identify Continuity
Problem 1. For each function, state the domain of continuity.
(a) $f(x) = \dfrac{2x + 1}{x^2 - 1}$
(b) $g(x) = \sqrt{3 - x}$
(c) $h(x) = \ln(x^2)$
Show answer
(a) Denominator: $x^2 - 1 = (x-1)(x+1)$, zero at $\pm 1$. Continuous on $(-\infty,-1) \cup (-1,1) \cup (1,\infty)$.
(b) $3 - x \geq 0$ when $x \leq 3$. Continuous on $(-\infty, 3]$.
(c) $\ln(x^2)$ requires $x^2 > 0$, i.e., $x \neq 0$. Continuous on $(-\infty, 0) \cup (0, \infty)$.
Problem 2. Evaluate each limit by identifying the appropriate continuity family.
(a) $\lim_{x \to 2} e^{3x - 5}$
(b) $\lim_{x \to \pi} \sin^2 x + \cos x$
(c) $\lim_{x \to 4} \ln\!\left(\sqrt{x} - 1\right)$
Show answer
(a) Exponential is continuous; $e^{3(2)-5} = e^1 = e$.
(b) $\sin^2\pi + \cos\pi = 0 + (-1) = -1$.
(c) Inner: $\sqrt{4} - 1 = 1 > 0$. $\ln$ is continuous at 1. Limit $= \ln(1) = 0$.
Level 2 -- Domain Analysis
Problem 3. Find all points where $f(x) = \dfrac{\sin x}{x^2 - \pi^2}$ is discontinuous. Classify each discontinuity (vertical asymptote or removable).
Show answer
Denominator: $x^2 - \pi^2 = (x-\pi)(x+\pi)$, zero at $x = \pm\pi$.
At $x = \pi$: numerator $= \sin\pi = 0$. Both vanish: $\frac{0}{0}$. Factor: $\frac{\sin x}{(x-\pi)(x+\pi)}$. Since $\lim_{x \to \pi}\frac{\sin x}{x - \pi}$ can be evaluated using $\lim_{u \to 0}\frac{\sin(\pi + u)}{u} = \lim_{u \to 0}\frac{-\sin u}{u} = -1$, and $\frac{1}{x+\pi} \to \frac{1}{2\pi}$, the limit is $\frac{-1}{2\pi}$. Removable discontinuity.
At $x = -\pi$: numerator $= \sin(-\pi) = 0$. Same type: $\frac{0}{0}$. Similarly, the limit exists. Removable discontinuity.
(Both are removable; if you redefine $f$ at these points to equal the limit, $f$ becomes continuous.)
Level 3 -- Composition and Families
Problem 4. Is $f(x) = e^{1/(x-2)}$ continuous everywhere? Identify and classify any discontinuities.
Show answer
The exponent $1/(x-2)$ is undefined at $x = 2$. For $x \neq 2$, it is continuous (rational function), and $e^{\text{anything finite}}$ is continuous. So $f$ is continuous for all $x \neq 2$.
At $x = 2$: as $x \to 2^+$, $1/(x-2) \to +\infty$, so $e^{1/(x-2)} \to +\infty$. As $x \to 2^-$, $1/(x-2) \to -\infty$, so $e^{1/(x-2)} \to 0$. The two one-sided limits differ (and the right-hand limit is infinite), so $f$ has an infinite discontinuity from the right at $x = 2$.
Common Misconceptions
every elementary function is continuous everywhere. Polynomials are continuous everywhere, but most other elementary families have restricted domains. Rational functions are continuous except where the denominator is zero. Logarithms require positive input. Even-root functions require non-negative input. Tangent and secant are undefined at isolated points. Continuity of a function family holds only on the natural domain of that family.
$\sqrt{x^2} = x$. For $x < 0$, $\sqrt{x^2} = |x| = -x \neq x$. In general, $\sqrt{x^2} = |x|$. Students sometimes cancel the square and root as inverse operations without checking the sign, which changes the domain or formula of the result.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
A function being continuous at $a$ means it has “no surprises” at $a$: what you predict from nearby values is exactly what you find at $a$.
The standard families are continuous on their domains because they are exactly the functions that mathematics has built to be smooth and predictable -- polynomials are smooth because they are built from sums and products of the smooth function $x$; trigonometric functions are smooth because they arise from the circle, which has no corners; exponentials and logarithms are smooth because they arise from the area under the hyperbola $y = 1/t$.
The domain exceptions (denominator zero, negative under even root, non-positive in log) are the only places these functions can fail. Everywhere else, plug in.
Connections
Within Calculus I (MATH161)
- Direct substitution: Direct substitution works because the function is built from continuous families.
- Continuity laws: The combination rules (sum, product, quotient, composition) extend the list of continuous functions indefinitely from the base families.
- Differentiability: Every differentiable function is continuous; the differentiable functions in Calculus I are also all built from these families.
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