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 [latex]f(x)[/latex] is continuous at [latex]L[/latex] and [latex]\underset{x\to a}{\lim}g(x)=L[/latex], then
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 [latex]\underset{x\to 0}{\lim} \cos x=1= \cos (0)[/latex]. Consequently, we know that [latex]f(x)= \cos x[/latex] is continuous at [latex]0[/latex]. In the next example we see how to combine this result with the composite function theorem.
Evaluate [latex]\underset{x\to \pi/2}{\lim}\cos(x-\dfrac{\pi }{2})[/latex].
Evaluate [latex]\underset{x\to \pi}{\lim}\sin(x-\pi)[/latex].
The proof of the next theorem uses the composite function theorem and the continuity of [latex]f(x)= \sin x[/latex] and [latex]g(x)= \cos x[/latex] at the point [latex]0[/latex] to demonstrate that trigonometric functions are continuous over their entire domains.
Proof
We begin by demonstrating that [latex]\cos x[/latex] is continuous at every real number. To do this, we must show that [latex]\underset{x\to a}{\lim}\cos x = \cos a[/latex] for all values of [latex]a[/latex].
[latex]\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}[/latex]
The proof that [latex]\sin x[/latex] is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of [latex]\sin x[/latex] and [latex]\cos x,[/latex] their continuity follows from the quotient limit law.
[latex]_\blacksquare[/latex]
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 [latex][a,b][/latex], where [latex]a[/latex] and [latex]b[/latex] 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 [latex]f[/latex] be continuous over a closed, bounded interval [latex][a,b][/latex]. If [latex]z[/latex] is any real number between [latex]f(a)[/latex] and [latex]f(b)[/latex], then there is a number [latex]c[/latex] in [latex][a,b][/latex] satisfying [latex]f(c)=z[/latex].
![A diagram illustrating the intermediate value theorem. There is a generic continuous curved function shown over the interval [a,b]. The points fa. and fb. are marked, and dotted lines are drawn from a, b, fa., and fb. to the points (a, fa.) and (b, fb.). A third point, c, is plotted between a and b. Since the function is continuous, there is a value for fc. along the curve, and a line is drawn from c to (c, fc.) and from (c, fc.) to fc., which is labeled as z on the y axis.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/2332/2018/01/11203518/CNX_Calc_Figure_02_04_007.jpg)
Show that [latex]f(x)=x- \cos x[/latex] has at least one zero.
If [latex]f(x)[/latex] is continuous over [latex][0,2], \, f(0)>0[/latex], and [latex]f(2)>0,[/latex] can we use the Intermediate Value Theorem to conclude that [latex]f(x)[/latex] has no zeros in the interval [latex][0,2][/latex]? Explain.
Show that [latex]f(x)=x^3-x^2-3x+1[/latex] has a zero over the interval [latex][0,1][/latex].