Continuity: Learn It 4

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Continuity: Learn It 4

Composite Function Theorem

The Composite Function Theorem helps expand our ability to compute limits, particularly demonstrating the continuity of trigonometric functions over their domains.

composite function theorem

If $f(x)$ is continuous at $L$ and $\underset{x\to a}{\lim}g(x)=L$ , then

lim_{x \to a} f (g (x)) = f (lim_{x \to a} g (x)) = f (L)

$\underset{x\to a}{\lim}f(g(x))=f(\underset{x\to a}{\lim}g(x))=f(L)$ .

This theorem allows us to demonstrate that the composition of functions is continuous if the inner function approaches a limit where the outer function is continuous.

Before we move on to the next example, recall that earlier, in the section on limit laws, we showed $\underset{x\to 0}{\lim} \cos x=1= \cos (0)$ . Consequently, we know that $f(x)= \cos x$ is continuous at $0$ . In the next example we see how to combine this result with the composite function theorem.

Evaluate $\underset{x\to \pi/2}{\lim}\cos(x-\dfrac{\pi }{2})$ .

The given function is a composite of $\cos x$ and $x-\frac{\pi}{2}$ . Since $\underset{x\to \pi/2}{\lim}(x-\frac{\pi}{2})=0$ and $\cos x$ is continuous at $0$ , we may apply the composite function theorem. Thus,

lim_{x \to π / 2} \cos (x - \frac{π}{2}) = \cos (lim_{x \to π / 2} (x - \frac{π}{2})) = \cos (0) = 1

$\underset{x\to \pi/2}{\lim}\cos(x-\frac{\pi}{2})= \cos (\underset{x\to \pi/2}{\lim}(x-\frac{\pi}{2}))= \cos (0)=1$ .

Evaluate $\underset{x\to \pi}{\lim}\sin(x-\pi)$ .

$f(x)= \sin x$ is continuous at $0$ .

$0$

Watch the following video to see examples of solving limits of composite functions.

The proof of the next theorem uses the composite function theorem and the continuity of $f(x)= \sin x$ and $g(x)= \cos x$ at the point $0$ to demonstrate that trigonometric functions are continuous over their entire domains.

Proof

We begin by demonstrating that $\cos x$ is continuous at every real number. To do this, we must show that $\underset{x\to a}{\lim}\cos x = \cos a$ for all values of $a$ .

$\begin{array}{lllll}\underset{x\to a}{\lim}\cos x & =\underset{x\to a}{\lim}\cos((x-a)+a) & & & \text{rewrite} \, x \, \text{as} \, x-a+a \, \text{and group} \, (x-a) \\ & =\underset{x\to a}{\lim}(\cos(x-a)\cos a - \sin(x-a)\sin a) & & & \text{apply the identity for the cosine of the sum of two angles} \\ & = \cos(\underset{x\to a}{\lim}(x-a)) \cos a - \sin(\underset{x\to a}{\lim}(x-a))\sin a & & & \underset{x\to a}{\lim}(x-a)=0, \, \text{and} \, \sin x \, \text{and} \, \cos x \, \text{are continuous at 0} \\ & = \cos(0)\cos a - \sin(0)\sin a & & & \text{evaluate cos(0) and sin(0) and simplify} \\ & =1 \cdot \cos a - 0 \cdot \sin a = \cos a \end{array}$

The proof that $\sin x$ is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of $\sin x$ and $\cos x,$ their continuity follows from the quotient limit law.

$_\blacksquare$

continuity of trigonometric functions

Trigonometric functions are continuous over their entire domains.

The Intermediate Value Theorem

Functions that are continuous over intervals of the form $[a,b]$ , where $a$ and $b$ are real numbers, exhibit many useful properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. The first of these theorems is the intermediate value theorem.

the intermediate value theorem

Let $f$ be continuous over a closed, bounded interval $[a,b]$ . If $z$ is any real number between $f(a)$ and $f(b)$ , then there is a number $c$ in $[a,b]$ satisfying $f(c)=z$ .

Show that $f(x)=x- \cos x$ has at least one zero.

Since $f(x)=x-\cos x$ is continuous over $(−\infty,+\infty)$ , it is continuous over any closed interval of the form $[a,b]$ . If you can find an interval $[a,b]$ such that $f(a)$ and $f(b)$ have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number $c$ in $(a,b)$ that satisfies $f(c)=0$ . Note that

f (0) = 0 - \cos (0) = - 1 < 0

$f(0)=0 - \cos (0)=-1<0$

and

f (\frac{π}{2}) = \frac{π}{2} - \cos \frac{π}{2} = \frac{π}{2} > 0

$f(\frac{\pi}{2})=\frac{\pi}{2} - \cos \frac{\pi}{2}=\frac{\pi}{2}>0$

Using the Intermediate Value Theorem, we can see that there must be a real number $c$ in $[0,\pi/2]$ that satisfies $f(c)=0$ . Therefore, $f(x)=x- \cos x$ has at least one zero.

If $f(x)$ is continuous over $[0,2], \, f(0)>0$ , and $f(2)>0,$ can we use the Intermediate Value Theorem to conclude that $f(x)$ has no zeros in the interval $[0,2]$ ? Explain.

No. The Intermediate Value Theorem only allows us to conclude that we can find a value between $f(0)$ and $f(2)$ ; it doesn’t allow us to conclude that we can’t find other values. To see this more clearly, consider the function $f(x)=(x-1)^2$ . It satisfies $f(0)=1>0, \, f(2)=1>0$ , and $f(1)=0$ .

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “2.4 Continuity” here (opens in new window).

Show that $f(x)=x^3-x^2-3x+1$ has a zero over the interval $[0,1]$ .

$f(0)=1>0, \, f(1)=-2<0$ ; $f(x)$ is continuous over $[0,1]$ . It must have a zero on this interval.