Chapter 5 · Continuity and Differentiability

Example 14: Every polynomial function is continuous

ExamplesContinuity · Section 5.2

The question

Show that every polynomial function is continuous.

Exam answer

Let $p(x) = a_0 + a_1x + \dots + a_nx^n$ ⁠, where $a_0, a_1, \dots, a_n$ are real constants. $p$ is defined for every real $x$ ⁠. Let $c$ be any real number.

Since $\lim_{x\to c} x = c$ ⁠, the product rule of limits gives $\lim_{x\to c} x^k = c^k$ for every natural number $k$ ⁠. So, by the sum and constant multiple rules of limits,

\begin{aligned} \lim_{x\to c} p(x) &= a_0 + a_1c + \dots + a_nc^n \\ &= p(c) \end{aligned}

$\therefore\ p$ is continuous at every real number, so every polynomial function is continuous. Hence proved.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

A polynomial is built from just two ingredients, constants and the variable $x$ , using only multiplication and addition. Limits respect both operations: the limit of a product is the product of the limits, and the limit of a sum is the sum of the limits. So taking $\lim_{x\to c}$ of a polynomial amounts to replacing every $x$ by $c$ ⁠, which gives exactly $p(c)$ ⁠. That equation, $\lim_{x\to c} p(x) = p(c)$ ⁠, is the definition of continuity at $c$ ⁠.

Given, to show, and the plan

Given: A polynomial function $p(x) = a_0 + a_1 x + \dots$ $+\, a_n x^n$ ⁠, with $n$ a natural number, real coefficients $a_i$ ⁠, and $a_n \ne 0$ ⁠.
To show: $p$ is continuous at every real number, so $p$ is a continuous function.
Plan: Domain. Fix an arbitrary real $c$ ⁠. Find the limits of constants and of $x$ ⁠; then of powers $x^k$ ⁠; then of each term $a_k x^k$ ⁠; then of the sum. Compare with $p(c)$ ⁠.

Basics you need

1What a polynomial function is

$p(x) = a_0 + a_1 x + a_2 x^2$ $+ \dots + a_n x^n$ ⁠, finitely many terms, each a real constant times a whole-number power of $x$ ⁠. Example: $x^3 + x^2 - 1$ (Example 8), with $a_0 = -1$ ⁠, $a_1 = 0$ ⁠, $a_2 = 1$ ⁠, $a_3 = 1$ ⁠.

2Algebra of limits (Class XI)

If $\lim_{x\to c} u(x)$ and $\lim_{x\to c} v(x)$ exist, then the limit of $u + v$ is the sum of the limits, the limit of $u \cdot v$ is the product of the limits, and the limit of $k\,u$ is $k$ times the limit. Basics: what a limit is

3Two starting limits

$\lim_{x\to c} k = k$ for a constant $k$ (Example 5) and $\lim_{x\to c} x = c$ (Example 6).

4Definitions 1 and 2

$p$ is continuous at $c$ when $\lim_{x\to c} p(x) = p(c)$ ⁠; $p$ is a continuous function when this holds at every point of its domain. Lesson: continuity at a point

Solution

The domain is all of $\R$

To compute $p(x)$ we only multiply and add real numbers, which is always possible. So $p$ is defined at every real number and its domain is $\R$ ⁠. Fix an arbitrary real number $c$ ⁠; we will show $\lim_{x\to c} p(x) = p(c)$ ⁠.

Definition 2 requires continuity at every point of the domain. Working with a letter $c$ that could be any real number proves it for all of them at once, instead of checking a few values.
The two building blocks

For a constant function and for the identity function:
$\lim_{x\to c} k = k, \qquad \lim_{x\to c} x = c.$
These are the simplest limits there are: a constant function has the value $k$ everywhere, and for the identity the output is the input, so as $x$ approaches $c$ the output approaches $c$ ⁠. The book proves both in Examples 5 and 6.

Example 5 (constants) and Example 6 (identity) are continuous everywhere.
Powers: $\lim_{x\to c} x^k = c^k$

For $k = 1$ this is Step 2. Suppose it is true for some power $k - 1$ ⁠, that is $\lim_{x\to c} x^{k-1} = c^{k-1}$ ⁠. Since $x^k = x \cdot x^{k-1}$ ⁠, the product rule gives
$\begin{aligned} \lim_{x\to c} x^k &= \Big(\lim_{x\to c} x\Big)\Big(\lim_{x\to c} x^{k-1}\Big) \\ &= c \cdot c^{k-1} = c^k. \end{aligned}$
Starting from $k = 1$ and repeating this step, it holds for $k = 2, 3, 4, \dots$ ⁠, so for every natural number $k$ ⁠.

The product rule of limits needs both factors to have limits. Each time we use it, the two factors are $x$ (limit $c$ ⁠, Step 2) and a lower power whose limit we already know. Climbing one power at a time reaches any power. For instance $\lim x^2 = c \cdot c$ ⁠, then $\lim x^3 = c \cdot c^2$ ⁠.

This "true for 1, and true for $k$ whenever true for $k - 1$ " argument is mathematical induction. Basics: what a limit is
Each term: $\lim_{x\to c} a_k x^k = a_k c^k$

Each $a_k$ is a constant, so by the constant-multiple rule and Step 3,
$\lim_{x\to c} a_k x^k = a_k \lim_{x\to c} x^k = a_k c^k.$
The constant term $a_0$ is handled by Step 2: $\lim_{x\to c} a_0 = a_0$ ⁠.

A constant factor can be taken outside a limit: if $x^k$ gets close to $c^k$ ⁠, then $a_k x^k$ gets close to $a_k c^k$ ⁠. This is the product rule with one factor constant.
Add the terms: $\lim_{x\to c} p(x) = p(c)$

$p(x)$ is the sum of the $n + 1$ terms $a_0, a_1 x, \dots, a_n x^n$ ⁠, each with a limit (Step 4). By the sum rule, applied $n$ times,
$\begin{aligned} \lim_{x\to c} p(x) &= a_0 + a_1 c + \dots + a_n c^n \\ &= p(c). \end{aligned}$
The sum rule is stated for two functions, but a sum of finitely many terms can be built two at a time, so it extends to any finite number of terms. The right-hand side is $p$ with every $x$ replaced by $c$ ⁠, which is precisely the number $p(c)$ ⁠.

The finiteness matters: a polynomial always has finitely many terms. Basics: what a limit is
Conclude with the definitions

Step 5 says $\lim_{x\to c} p(x) = p(c)$ ⁠. By Definition 1, $p$ is continuous at $c$ ⁠. Since $c$ was an arbitrary real number and the domain is $\R$ ⁠, $p$ is continuous at every point of its domain, and by Definition 2 $p$ is a continuous function.

This is the textbook's argument with each limit justified. The book writes $\lim_{x\to c} p(x) = p(c)$ in one line; Steps 2 to 5 show why that line is true.

Constant polynomials ( $p(x) = a_0$ ⁠) are not covered by the book's description with $a_n \ne 0$ and $n \ge 1$ ⁠, but they are continuous by Example 5, so the result holds for every polynomial. Lesson: continuity at a point
Another route: Theorem 1

In Section 5.2.1, after Example 15, the book states Theorem 1 (sums, differences, products and quotients of continuous functions are continuous). It proves part (1) and leaves the other parts, including part (3) used here, to the reader; they follow from the product and quotient rules of limits just as part (1) follows from the sum rule. With Theorem 1 the result is immediate: $x$ and constants are continuous, so $x^k = x \cdot x \cdots x$ is continuous by part (3) applied repeatedly, $a_k x^k$ is continuous by part (3) again (Remark (i)), and the sum of the terms is continuous by part (1).

Theorem 1 is just the algebra of limits, restated for continuous functions. The two routes are the same argument in different words, which is why the book's later examples (rational functions, Example 16) can lean on this one.

Lesson: algebra of continuous functions

Result

For every polynomial $p$ and every real number $c$ ⁠, $\lim_{x\to c} p(x) = p(c)$ ⁠. Hence every polynomial function is continuous (at every real number). This agrees with the textbook's solution.

Check it with numbers

Take $p(x) = x^3 + x^2 - 1$ (Example 8) and $c = 2$ ⁠: $p(2) = 8 + 4 - 1 = 11$ ⁠. Nearby,

p(1.999) = 10.984006999,

p(2.001) = 11.016007001,

both within $0.02$ of 11. With $x = 2 \pm 0.000001$ the outputs differ from 11 by about $0.000016$ ⁠: the closer the input, the closer the output, as continuity predicts.

Common mistakes

"Just substitute" without a reason. Substituting $x = c$ gives the limit only because the function is continuous; for a piecewise function such as Example 11, substituting into a formula can give the wrong limit. Here Steps 2 to 5 are the reason.
Checking a few polynomials. Verifying $x^2$ and $x^3 + x^2 - 1$ does not prove the statement for every polynomial. Work with general coefficients $a_0, \dots, a_n$ ⁠.
Forgetting the domain. A polynomial is defined at every real number, which is why "continuous" here means "continuous at every real number".

What to take away

For every polynomial, $\lim_{x\to c} p(x) = p(c)$ ⁠: limits of polynomials are found by substitution.
The proof grows the result from the simplest functions (constants and $x$ ⁠) using the sum and product rules of limits.
This result is used constantly later: in rational functions (Example 16), in $\sin(x^2)$ (Example 19) and in $|1 - x + |x||$ (Example 20).

Every fact used, and where it comes from

Fact	Used in	Source
Polynomial function $a_0 + a_1x + \dots + a_nx^n$	Steps 1, 5	Example 14, Section 5.2; Class XI
$\lim_{x\to c} k = k$ ⁠, $\lim_{x\to c} x = c$	Steps 2, 3, 4	Examples 5 and 6, Section 5.2
Sum, product and constant-multiple rules of limits	Steps 3, 4, 5	Algebra of limits, Class XI (Limits and Derivatives) (Basics: limits)
Continuity at a point; continuous function	Step 6	Definitions 1 and 2, Section 5.2 (Lesson: continuity at a point)
Sums and products of continuous functions are continuous	Step 7	Theorem 1 and Remark (i), Section 5.2.1 (Lesson: algebra of continuous functions)

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Example 14: Every polynomial function is continuous

Basics you need

Solution

The domain is all of R\RR

The two building blocks

Powers: lim⁡x→cxk=ck\lim_{x\to c} x^k = c^klimx→c​xk=ck

Each term: lim⁡x→cakxk=akck\lim_{x\to c} a_k x^k = a_k c^klimx→c​ak​xk=ak​ck

Add the terms: lim⁡x→cp(x)=p(c)\lim_{x\to c} p(x) = p(c)limx→c​p(x)=p(c)

Conclude with the definitions

Another route: Theorem 1

Every fact used, and where it comes from

The domain is all of $\R$

Powers: $\lim_{x\to c} x^k = c^k$

Each term: $\lim_{x\to c} a_k x^k = a_k c^k$

Add the terms: $\lim_{x\to c} p(x) = p(c)$