Chapter 5 · Continuity and Differentiability

Example 8: Continuity of $f(x)=x^3+x^2-1$

ExamplesContinuous functions · Section 5.2

The question

Discuss the continuity of the function $f$ given by $f(x) = x^3 + x^2 - 1$ ⁠.

Exam answer

$f$ is a polynomial, defined for every real $x$ ⁠. Let $c$ be any real number.

\begin{aligned} \lim_{x\to c} f(x) &= \lim_{x\to c}(x^3 + x^2 - 1) \\ &= c^3 + c^2 - 1 = f(c) \end{aligned}

$\therefore\ f$ is continuous at every real number, so $f$ is a continuous function.

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

The one idea this problem is built on

A polynomial is built from $x$ and constants using only addition and multiplication, and limits respect addition and multiplication. So the limit of $x^3 + x^2 - 1$ at any point $c$ is found by putting $x = c$ ⁠, which is exactly the value $f(c)$ ⁠. Limit equals value at every $c$ ⁠: the function is continuous everywhere.

Given, to find, and the plan

Given: $f(x) = x^3 + x^2 - 1$ for every real $x$ ⁠.
To find: The points where $f$ is continuous, and whether $f$ is a continuous function.
Plan: Take an arbitrary real $c$ ⁠. Find $f(c)$ ⁠. Compute the left-hand and right-hand limits at $c$ with the algebra of limits, confirm them by measuring $|f(x) - f(c)|$ ⁠, and compare (Definition 1).

Basics you need

1Algebra of limits (Class XI)

If $\lim g = A$ and $\lim h = B$ (at the same point, from the same side), then $\lim (g + h) = A + B$ ⁠, $\lim (g - h) = A - B$ and $\lim (gh) = AB$ ⁠. Also $\lim_{x\to c} x = c$ and $\lim_{x\to c} k = k$ ⁠. Basics: what a limit is

2Powers as repeated products

$x^2 = x\cdot x$ and $x^3 = x\cdot x\cdot x$ ⁠, so the product rule gives $\lim_{x\to c} x^2 = c^2$ and $\lim_{x\to c} x^3 = c^3$ ⁠.

3One-sided limits

The limit at $c$ exists when the left-hand and right-hand limits exist and agree. Basics: left-hand and right-hand limits

4Definitions 1 and 2

Continuous at $c$ (in the domain): left-hand limit = right-hand limit = $f(c)$ ⁠. Continuous function: continuous at every point of its domain. Lesson: continuity at a point

Solution

Domain, and the value at an arbitrary point

The polynomial $x^3 + x^2 - 1$ can be evaluated at every real number, so the domain is $\R$ ⁠. Let $c$ be any real number. Then
$f(c) = c^3 + c^2 - 1.$
"Discuss the continuity" means: find every point where $f$ is (or is not) continuous. Working at an arbitrary $c$ handles all real numbers at once.

Only additions and multiplications are involved, so there is no division by zero or root of a negative number to exclude.
Left-hand and right-hand limits by the algebra of limits

The same formula is used on both sides of $c$ ⁠. Using Basics 1 and 2, from either side,
$\lim_{x\to c^\pm}(x^3 + x^2 - 1) = c^3 + c^2 - 1.$
In detail: $\lim x^3 = c\cdot c\cdot c = c^3$ ⁠, $\lim x^2 = c\cdot c = c^2$ and $\lim 1 = 1$ ⁠; then the sum and difference rules combine them.

Here $c^\pm$ is shorthand for " $c^-$ and also $c^+$ ": the computation is identical for the two sides, because the formula does not change at $c$ ⁠. So $\lim_{x\to c} f(x) = c^3 + c^2 - 1$ ⁠.

The limit rules hold for one-sided limits as well as for two-sided ones. Basics: one-sided limits
Confirm it: how far is $f(x)$ from $f(c)$ ⁠?

Subtract and group the matching powers:
$\begin{aligned} f(x) - f(c) &= (x^3 - c^3) + (x^2 - c^2) \\ &= (x - c)(x^2 + xc + c^2) \\ &\quad + (x - c)(x + c) \\ &= (x - c)(x^2 + xc + c^2 + x + c). \end{aligned}$
When $|x - c| < 1$ ⁠, we have $|x| < |c| + 1$ ⁠, so the second bracket is at most $3(|c|+1)^2 + 2(|c|+1)$ in size: a fixed number, call it $M$ ⁠. Then
$|f(x) - f(c)| \le M\,|x - c|,$
which can be made as small as we like by taking $x$ close enough to $c$ ⁠.

The algebra of limits gives the answer; this step shows why it is right for this polynomial. The factor $x - c$ shrinks to 0 while the other factor stays bounded, so the product shrinks to 0. That is exactly " $f(x)$ can be made as close to $f(c)$ as we like".

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ and $a^2 - b^2 = (a - b)(a + b)$ ⁠. The bound uses $|x^2| , |xc| , |c^2| < (|c|+1)^2$ and $|x| , |c| < |c| + 1$ ⁠. Basics: what a limit is
Compare at $c$
$\lim_{x\to c^-} f(x) = \lim_{x\to c^+} f(x) = c^3 + c^2 - 1 = f(c).$
All three agree, so $f$ is continuous at $c$ (Definition 1). Since $c$ was an arbitrary real number, $f$ is continuous at every real number, and therefore $f$ is a continuous function (Definition 2).

This is the textbook's argument: $\lim_{x\to c} f(x) = f(c)$ at every real $c$ ⁠. Because the domain is all of $\R$ ⁠, "continuous at every real number" and "continuous function" mean the same thing here.

Lesson: continuity at a point

The cubic

y = x^3 + x^2 - 1

(horizontal unit 92 px, vertical unit 40 px): an unbroken curve. At the sample point

c = 1

the curve arrives at height

f(1) = 1

from both sides, and the same happens at every other

c

⁠.

Result

For every real $c$ ⁠, $\lim_{x\to c} f(x) = c^3 + c^2 - 1$ ⁠, which is $f(c)$ ⁠. So $f(x) = x^3 + x^2 - 1$ is continuous at every real number; it is a continuous function. This agrees with the textbook.

Check it with numbers

At $c = 1$ ⁠, $f(1) = 1 + 1 - 1 = 1$ ⁠. Nearby values:

$x$	$f(x)$	side of 1
$0.9$	$0.539$	left
$0.99$	$0.950399$	left
$0.999$	$0.995003999$	left
$1.001$	$1.005004001$	right
$1.01$	$1.050401$	right
$1.1$	$1.541$	right

At distance $0.001$ from 1, $f(x)$ is about $0.005$ from 1. That matches Step 3: the second bracket at $x = c = 1$ is $1 + 1 + 1 + 1 + 1 = 5$ ⁠, so $f(x) - f(1) \approx 5(x - 1)$ near 1.

Common mistakes

Checking a handful of points. The question is about every real number, so argue at an arbitrary $c$ ⁠.
Treating "put $x = c$ " as the definition of a limit. It is a consequence of the limit rules for polynomials; it fails for functions like the one in Example 4.
Dropping the constant. $\lim_{x\to c}(-1) = -1$ ⁠, so $f(c)$ ends in $-1$ ⁠, not 0.

What to take away

For a polynomial, the limit at any $c$ equals the value at $c$ ⁠, so every polynomial is continuous on $\R$ ⁠. The general proof is Example 14, using Theorem 1 of Section 5.2.1.
The factor $x - c$ appears in $p(x) - p(c)$ for every polynomial $p$ ⁠; it is what forces $p(x)$ close to $p(c)$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Limits of sums, differences and products; $\lim x = c$ ⁠, $\lim k = k$	Step 2	Algebra of limits, Class XI (Limits and Derivatives) (Basics: limits)
$a^3 - b^3 = (a-b)(a^2+ab+b^2)$ ⁠; $a^2 - b^2 = (a-b)(a+b)$	Step 3	Algebraic identities
$\|a + b\| \le \|a\| + \|b\|$ ⁠; $\|ab\| = \|a\|\|b\|$	Step 3	Properties of absolute value
Continuous at $c$ iff LHL = RHL = $f(c)$	Step 4	Definition 1, Section 5.2 (Lesson: continuity at a point)
Continuous function	Step 4	Definition 2, Section 5.2

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Example 8: Continuity of f(x)=x3+x2−1f(x)=x^3+x^2-1f(x)=x3+x2−1

Basics you need

Solution

Domain, and the value at an arbitrary point

Left-hand and right-hand limits by the algebra of limits

Confirm it: how far is f(x)f(x)f(x) from f(c)f(c)f(c)⁠?

Compare at ccc

Every fact used, and where it comes from

Example 8: Continuity of $f(x)=x^3+x^2-1$

Confirm it: how far is $f(x)$ from $f(c)$ ⁠?

Compare at $c$