Now that we have explored the concept of continuity at a point, we extend that idea to continuity over an interval. As we develop this idea for different types of intervals, it may be useful to keep in mind the intuitive idea that a function is continuous over an interval if we can use a pencil to trace the function between any two points in the interval without lifting the pencil from the paper. In preparation for defining continuity on an interval, we begin by looking at the definition of what it means for a function to be continuous from the right at a point and continuous from the left at a point.

A function is continuous over an open interval if it is continuous at every point in the interval. A function $f(x)$ is continuous over a closed interval of the form $[a,b]$ if it is continuous at every point in $(a,b)$ and is continuous from the right at $a$ and is continuous from the left at $b$. Analogously, a function $f(x)$ is continuous over an interval of the form $(a,b]$ if it is continuous over $(a,b)$ and is continuous from the left at $b$. Continuity over other types of intervals are defined in a similar fashion.

Requiring that $\underset{x\to a^+}{\lim}f(x)=f(a)$ and $\underset{x\to b^-}{\lim}f(x)=f(b)$ ensures that we can trace the graph of the function from the point $(a,f(a))$ to the point $(b,f(b))$ without lifting the pencil. If, for example, $\underset{x\to a^+}{\lim}f(x)\ne f(a)$, we would need to lift our pencil to jump from $f(a)$ to the graph of the rest of the function over $(a,b]$.

The Composite Function Theorem allows us to expand our ability to compute limits. In particular, this theorem ultimately allows us to demonstrate that trigonometric functions are continuous over their domains.

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

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.

The proof of the next theorem uses the composite function theorem as well as the continuity of $f(x)= \sin x$ and $g(x)= \cos x$ at the point 0 to show 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.

$_\blacksquare$

As you can see, the composite function theorem is invaluable in demonstrating the continuity of trigonometric functions. As we continue our study of calculus, we revisit this theorem many times.

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

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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 $f(c)=z$. (See Figure 7).

Figure 7. There is a number $c \in [a,b]$ that satisfies $f(c)=z$.

Example: Application of the Intermediate Value Theorem

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

Show Solution

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

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

and

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

 

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.

Watch the following video to see the worked solution to Example: When Can You Apply the Intermediate Value Theorem?.