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Requiring that $\underset{x\to {a}^{+}}{\text{lim}}f\left(x\right)=f\left(a\right)$ and $\underset{x\to {b}^{-}}{\text{lim}}f\left(x\right)=f\left(b\right)$ ensures that we can trace the graph of the function from the point $\left(a,f\left(a\right)\right)$ to the point $\left(b,f\left(b\right)\right)$ without lifting the pencil. If, for example, $\underset{x\to {a}^{+}}{\text{lim}}f\left(x\right)\ne f\left(a\right),$ we would need to lift our pencil to jump from $f\left(a\right)$ to the graph of the rest of the function over $\left(a,b\right].$
State the interval(s) over which the function $f\left(x\right)=\frac{x-1}{{x}^{2}+2x}$ is continuous.
Since $f\left(x\right)=\frac{x-1}{{x}^{2}+2x}$ is a rational function, it is continuous at every point in its domain. The domain of $f\left(x\right)$ is the set $\left(\text{\u2212}\infty ,\mathrm{-2}\right)\cup \left(\mathrm{-2},0\right)\cup \left(0,\text{+}\infty \right).$ Thus, $f\left(x\right)$ is continuous over each of the intervals $\left(\text{\u2212}\infty ,\mathrm{-2}\right),\left(\mathrm{-2},0\right),$ and $\left(0,\text{+}\infty \right).$
State the interval(s) over which the function $f(x)=\sqrt{4-{x}^{2}}$ is continuous.
From the limit laws, we know that $\underset{x\to a}{\text{lim}}\sqrt[]{4-{x}^{2}}=\sqrt{4-{a}^{2}}$ for all values of a in $\left(\mathrm{-2},2\right).$ We also know that $\underset{x\to {\mathrm{-2}}^{+}}{\text{lim}}\sqrt{4-{x}^{2}}=0$ exists and $\underset{x\to {2}^{-}}{\text{lim}}\sqrt{4-{x}^{2}}=0$ exists. Therefore, $f\left(x\right)$ is continuous over the interval $\left[\mathrm{-2},2\right].$
State the interval(s) over which the function $f\left(x\right)=\sqrt{x+3}$ is continuous.
$\left[\mathrm{-3},\text{+}\infty \right)$
The [link] 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.
If $f\left(x\right)$ is continuous at L and $\underset{x\to a}{\text{lim}}g\left(x\right)=L,$ then
Before we move on to [link] , recall that earlier, in the section on limit laws, we showed $\underset{x\to 0}{\text{lim}}\text{cos}\phantom{\rule{0.1em}{0ex}}x=1=\text{cos}\phantom{\rule{0.1em}{0ex}}\left(0\right).$ Consequently, we know that $f\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x$ is continuous at 0. In [link] we see how to combine this result with the composite function theorem.
Evaluate $\underset{x\to \pi \text{/}2}{\text{lim}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(x-\frac{\pi}{2}\right).$
The given function is a composite of $\text{cos}\phantom{\rule{0.1em}{0ex}}x$ and $x-\frac{\pi}{2}.$ Since $\underset{x\to \pi \text{/}2}{\text{lim}}\left(x-\frac{\pi}{2}\right)=0$ and $\text{cos}\phantom{\rule{0.1em}{0ex}}x$ is continuous at 0, we may apply the composite function theorem. Thus,
Evaluate $\underset{x\to \pi}{\text{lim}}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(x-\pi \right).$
0
The proof of the next theorem uses the composite function theorem as well as the continuity of $f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$ and $g\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x$ at the point 0 to show that trigonometric functions are continuous over their entire domains.
Trigonometric functions are continuous over their entire domains.
We begin by demonstrating that $\text{cos}\phantom{\rule{0.1em}{0ex}}x$ is continuous at every real number. To do this, we must show that $\underset{x\to a}{\text{lim}}\text{cos}\phantom{\rule{0.1em}{0ex}}x=\text{cos}\phantom{\rule{0.1em}{0ex}}a$ for all values of a .
$\begin{array}{ccccc}\underset{x\to a}{\text{lim}}\text{cos}\phantom{\rule{0.1em}{0ex}}x\hfill & =\underset{x\to a}{\text{lim}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(\left(x-a\right)+a\right)\hfill & & & \text{rewrite}\phantom{\rule{0.2em}{0ex}}x=x-a+a\hfill \\ & =\underset{x\to a}{\text{lim}}\left(\text{cos}\phantom{\rule{0.1em}{0ex}}\left(x-a\right)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}a-\text{sin}\phantom{\rule{0.1em}{0ex}}\left(x-a\right)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}a\right)\hfill & & & \text{apply the identity for the cosine of the sum of two angles}\hfill \\ & =\text{cos}\phantom{\rule{0.1em}{0ex}}\left(\underset{x\to a}{\text{lim}}\left(x-a\right)\right)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}a-\text{sin}\phantom{\rule{0.1em}{0ex}}\left(\underset{x\to a}{\text{lim}}\left(x-a\right)\right)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}a\hfill & & & \underset{x\to a}{\text{lim}}\left(x-a\right)=0,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{are continuous at 0}\hfill \\ & =\text{cos}\phantom{\rule{0.1em}{0ex}}\left(0\right)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}a-\text{sin}\phantom{\rule{0.1em}{0ex}}\left(0\right)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}a\hfill & & & \text{evaluate cos(0) and sin(0) and simplify}\hfill \\ & =1\xb7\text{cos}\phantom{\rule{0.1em}{0ex}}a-0\xb7\text{sin}\phantom{\rule{0.1em}{0ex}}a=\text{cos}\phantom{\rule{0.1em}{0ex}}a.\hfill \end{array}$
The proof that $\text{sin}\phantom{\rule{0.1em}{0ex}}x$ is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of $\text{sin}\phantom{\rule{0.1em}{0ex}}x$ and $\text{cos}\phantom{\rule{0.1em}{0ex}}x,$ their continuity follows from the quotient limit law.
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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.
Functions that are continuous over intervals of the form $\left[a,b\right],$ 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 .
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