Chapter 5 · Continuity and Differentiability

Example 19: $\sin(x^2)$ is continuous: a composite of continuous functions

ExamplesAlgebra of continuous functions · Section 5.2.1

The question

Show that the function defined by $f(x) = \sin(x^2)$ is a continuous function.

Exam answer

Let $h(x) = x^2$ and $g(x) = \sin x$ ⁠. Then

g(h(x)) = g(x^2) = \sin(x^2) = f(x)

$h$ is a polynomial, so it is continuous, and $g$ (the sine function) is continuous at every real number.

A composite of continuous functions is continuous, so $f(x) = g(h(x))$ is continuous at every real number.

$\therefore\ f(x) = \sin(x^2)$ is a continuous function. Hence proved.

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

The one idea this problem is built on

To compute $\sin(x^2)$ we do two things in turn: first square, then take the sine. So $f$ is a composite of two functions we already know are continuous. Theorem 2 says a composite is continuous at $c$ when the inner function is continuous at $c$ and the outer function is continuous at the inner value. Here both are continuous everywhere, so $f$ is continuous everywhere.

Given, to show, and the plan

Given: $f(x) = \sin(x^2)$ for real $x$ ⁠.
To show: $f$ is continuous at every point of its domain.
Plan: Domain. Write $f = g \circ h$ with inner $h(x) = x^2$ and outer $g(x) = \sin x$ ⁠, checking the order. Show $h$ is continuous at $c$ and $g$ at $h(c)$ ⁠. Apply Theorem 2.

Basics you need

1Composite functions

$(g \circ h)(x) = g(h(x))$ ⁠: apply $h$ first, then $g$ to the result. The function applied first is the inner one. Basics: composite functions

2Theorem 2 (continuity of composites)

If $f \circ g$ is defined at $c$ ⁠, $g$ is continuous at $c$ ⁠, and $f$ is continuous at $g(c)$ ⁠, then $f \circ g$ is continuous at $c$ ⁠. The book states it without proof. Lesson: algebra of continuous functions

3Two known continuous functions

$x^2$ is a polynomial, continuous at every real number (Example 14); $\sin x$ is continuous at every real number (Example 17).

4Definition 2

A function is continuous if it is continuous at every point of its domain. Lesson: continuity at a point

Solution

The domain is all of $\R$

For every real $x$ ⁠, $x^2$ is a real number, and sine is defined for every real number. So $\sin(x^2)$ is defined for every real $x$ ⁠: the domain is $\R$ ⁠.

Theorem 2 needs the composite to be defined at the point $c$ ⁠, and Definition 2 needs continuity at every point of the domain. Both are settled by knowing the domain is all of $\R$ ⁠.
Split $f$ into inner and outer functions

Let $h(x) = x^2$ (inner) and $g(x) = \sin x$ (outer). Then
$(g \circ h)(x) = g(h(x)) = g(x^2) = \sin(x^2) = f(x).$
So $f = g \circ h$ ⁠. The order matters: the other composite is different,
$(h \circ g)(x) = h(\sin x) = \sin^2 x.$
For example at $x = 1$ ⁠: $\sin(1^2) \approx 0.841471$ ⁠, but $\sin^2 1 \approx 0.708073$ ⁠.

To find the inner function, ask "what is done to $x$ first?" In $\sin(x^2)$ we square first, so squaring is inner and sine is outer. Getting this backwards would describe $\sin^2 x$ ⁠, a different function.

Basics: composite functions
The inner function is continuous at $c$

Fix any real $c$ ⁠. $h(x) = x^2$ is a polynomial, so by Example 14
$\lim_{x\to c} x^2 = c^2 = h(c).$
Theorem 2's first condition: the inner function must be continuous at the point $c$ itself.
The outer function is continuous at $h(c)$

$h(c) = c^2$ is a real number, and $g(x) = \sin x$ is continuous at every real number (Example 17), so in particular at $c^2$ ⁠:
$\lim_{u\to c^2} \sin u = \sin(c^2).$
Theorem 2's second condition is about the outer function at the point $h(c)$ ⁠, not at $c$ ⁠. Here this costs nothing, because sine is continuous everywhere, but in other problems this is where care is needed.

Example 17: the sine function is continuous.
Apply Theorem 2

$g \circ h$ is defined at $c$ ⁠, $h$ is continuous at $c$ ⁠, and $g$ is continuous at $h(c)$ ⁠. By Theorem 2, $f = g \circ h$ is continuous at $c$ ⁠:
$\lim_{x\to c}\sin(x^2) = \sin(c^2) = f(c).$
Since $c$ was any real number, $f$ is continuous at every point of its domain $\R$ ⁠, and by Definition 2 $f(x) = \sin(x^2)$ is a continuous function.

The idea behind Theorem 2: as $x \to c$ ⁠, the inner output $u = x^2$ approaches $c^2$ (Step 3); and as $u \to c^2$ ⁠, the outer output $\sin u$ approaches $\sin(c^2)$ (Step 4). Chaining the two approaches gives $\sin(x^2) \to \sin(c^2)$ ⁠.

Theorem 2, Section 5.2.1. Lesson: algebra of continuous functions

Result

$f = g \circ h$ with $h(x) = x^2$ and $g(x) = \sin x$ ⁠, both continuous everywhere, so by Theorem 2 $\lim_{x\to c}\sin(x^2) = \sin(c^2)$ for every real $c$ ⁠. Hence $f(x) = \sin(x^2)$ is a continuous function. This agrees with the textbook's solution.

Check it with numbers

At $c = 1.5$ ⁠: $c^2 = 2.25$ and $f(1.5) = \sin 2.25 \approx 0.778073$ ⁠. Nearby,

f(1.499) = \sin 2.247001 \approx 0.779954,

f(1.501) = \sin 2.253001 \approx 0.776185,

both within $0.002$ of $f(1.5)$ ⁠: small changes in $x$ give small changes in $x^2$ ⁠, which give small changes in $\sin(x^2)$ ⁠.

Common mistakes

Mixing up the order. $\sin(x^2) = (g \circ h)(x)$ ⁠, not $(h \circ g)(x)$ ⁠, which is $\sin^2 x$ ⁠.
Checking the outer function at the wrong point. Theorem 2 needs $g$ continuous at $h(c) = c^2$ ⁠, not at $c$ ⁠.
Using Theorem 1 instead. $\sin(x^2)$ is not a sum, product or quotient of $\sin x$ and $x^2$ ⁠; it is a composite, so Theorem 2 is the right tool.

What to take away

A composite of continuous functions is continuous (Theorem 2): find the inner and outer functions, then check each where it is needed.
Inner function = the operation done to $x$ first.
The same method handles $|1 - x + |x||$ in Example 20, and in Section 5.3 the chain rule differentiates such composites.

Every fact used, and where it comes from

Fact	Used in	Source
$(g \circ h)(x) = g(h(x))$	Step 2	Composition of functions, Chapter 1 (Basics: composite functions)
$x^2$ is continuous (a polynomial)	Step 3	Example 14, Section 5.2
$\sin x$ is continuous	Step 4	Example 17, Section 5.2.1
Composite of continuous functions is continuous	Step 5	Theorem 2, Section 5.2.1 (Lesson: algebra of continuous functions)
Continuous function	Step 5	Definition 2, Section 5.2 (Lesson: continuity at a point)

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Example 19: sin⁡(x2)\sin(x^2)sin(x2) is continuous: a composite of continuous functions

Basics you need

Solution

The domain is all of R\RR

Split fff into inner and outer functions

The inner function is continuous at ccc

The outer function is continuous at h(c)h(c)h(c)

Apply Theorem 2

Every fact used, and where it comes from

Example 19: $\sin(x^2)$ is continuous: a composite of continuous functions

The domain is all of $\R$

Split $f$ into inner and outer functions

The inner function is continuous at $c$

The outer function is continuous at $h(c)$