### Definitions of Continuity and Discontinuity

Consider a function $$f$$ whose graph is shown in the following figure. We can intuitively say that $$f$$ is discontinuous at $$x=-3,-1,3$$, and $$5$$ and is continuous at any other points; we don’t have to lift the pen to draw the graph of $$f$$ except when $$x=-3,-1,3,5$$.

• The function is discontinuous at $$x=-3$$ because it is not defined there.

• The function has a jump discontinuity at $$x=-1$$; the left-hand limit is not equal to the right-hand limit. Because $$\lim_{x\to-1^{-}}f(x)=f(-1)$$, we say the function is continuous from the left.

• The function is not continuous at $$x=3$$ because $$f(3)$$ is not equal to the limit of the function as $$x\to3$$. $\lim_{x\to3}f(x)=-1\neq f(3)=5$

• The function is discontinuous at $$x=5$$ because $\lim_{x\to5^{-}}f(x)=3\neq f(5)=1.$ However, we intuitively say the function continuous at $$x=-5$$ because $\lim_{x\to-5^{+}}f(x)=f(5)=1.$ Because $$x=\pm5$$ are endpoints of the domain of $$f$$, only one-sided limits exist.

Here is the formal definition of continuity.

Definition 1: The function $$f$$ is continuous at $$a$$ if $\lim_{x\to a}f(x)=f(a)$

It follows from the above definition that when the function $$f$$ is continuous at $$a$$ then

1. $$f$$ is defined at $$a$$.
2. $$\lim_{x\to a}f(x)$$ exits (which requires that $$f$$ to be defined on some open interval containing $$a$$).
3. $$\lim_{x\to a}f(x)=f(a)$$
• When $$f$$ is not continuous at $$a$$ we say $$f$$ is discontinuous at $$a$$ or has a discontinuity at $$a$$.

### Left and Right Continuities

In a similar fashion, we can define left and right continuities.

Definition 2: The function $$f$$ is left-continuous at $$a$$ (or continuous from the left) if $\lim_{x\to a^{-}}f(x)=f(a)$ The function is right-continuous at $$a$$ (or continuous from the right) if $\lim_{x\to a^{+}}f(x)=f(a).$

• It follows from the definitions that the function $$f$$ is continuous at $$x=a$$ if and only if it is left-continuous and right-continuous at the point $$a$$.

### Continuity on an Interval

• If a function is continuous at every $$x$$ in an open interval $$(a,b)$$, we say it is continuous over $$(a,b)$$.
• We say a function is continuous over a closed interval $$[a,b]$$ if it is continuous over the open interval $$(a,b)$$ and is left-continuous at $$a$$ and right-continuous at $$b$$.
Example 1
Show that $$f(x)=3-\sqrt{4-x^{2}}$$ is continuous on $$[-2,2]$$.

Solution

For $$-2<a<2$$, we have

\begin{aligned} \lim_{x\to a}f(x) & =\lim_{x\to a}(3-\sqrt{4-x^{2}})\\ & =\lim_{x\to a}3-\lim_{x\to a}\sqrt{4-x^{2}} & {\small(\text{by Difference Rule in Sec. 4.4)}}\\ & =3-\sqrt{\lim_{x\to a}(4-x^{2})} & {\small( \text{by Root Rule})}\\ & =3-\sqrt{4-a^{2}}& {\small ((4-x^2) \text{ is a Polynomial})}\\ & =f(a) \end{aligned}

Similarly, we can show $\lim_{x\to-2^{+}}f(x)=3=f(-2),\quad\lim_{x\to2^{-}}f(x)=3=f(2)$ The graph of $$f$$ is shown in the following figure.

### Types of Discontinuity

1. Removable Discontinuity

We say that $$f$$ has a removable discontinuity at $$x=a$$ if $\lim_{x\to a}f(x)=L,$ and either $$f$$ is not defined at $$a$$ or $$f(a)\neq L$$. We can make $$f(x)$$ continuous at $$x=a$$, if $$L$$ is assumed as the value of $$f(a).$$

#### Read More on Removable Discontinuity

For example, the function $f(x)=\frac{x^{2}-4}{x-2}$ is not defined for $$x=2$$ (because there would be division by zero), but for every other value of $$x$$, $f(x)=\frac{(x-2)(x+2)}{x-2}=x+2.$ Because $\lim_{x\to2}(x+2)=4,$ we obtain $\lim_{x\to2}\frac{x^{2}-4}{x-2}=4.$ Although the function is not defined when $$x=2$$, if we arbitrarily assign $$f(x)$$ the value 4 when $$x=2$$ (that is, $$f(2)=4$$), then it becomes continuous at $$x=2$$ (see Figure 3).

 (a) $f(x)=\frac{x^{2}-4}{x-2}$ is discontinuous at $x=2$ (b) The discontinuity of $f$ at $x=2$ is removed if we define $f(x)=\begin{cases} \frac{x^{2}-4}{x-2} & x\neq2\\ 4 & x=2 \end{cases}$.

Figure 3

2. Jump Discontinuity

We say $$f$$ has a jump discontinuity at $$x=a$$ if the right and left limits exist but have different values $\lim_{x\to a^{-}}f(x)\neq\lim_{x\to a^{+}}f(x).$

#### Read More on Jump Discontinuity

For example, the Heaviside step function $$H(x)$$ is defined as $H(x)=\begin{cases} 1 & \text{if }x\geq0\\ 0 & \text{if }x<0 \end{cases}$ This function has a jump discontinuity at $$x=0$$ because $\lim_{x\to0^{-}}H(x)=0\neq\lim_{x\to0^{+}}H(x)=1.$ We note that $$H(x)$$ is right continuous at $$x=0$$ because $$H(0)=\lim_{x\to0^{+}}H(x)=1$$. (See Figure 4).

 Figure 4: Graph of $$H(x)=\begin{cases} 1 & \text{if }x\geq0\\ 0 & \text{if }x<0 \end{cases}$$

The greatest integer function (also known as the floor function) $$y=\left\lfloor x\right\rfloor$$ (or $y=[\![ x]\!]$) has a jump discontinuity at every integer (see Figure 5). For example, $\lim_{x\to2^{-}}\left\lfloor x\right\rfloor =1\neq\lim_{x\to2^{+}}\left\lfloor x\right\rfloor =2.$

 Figure 5: Graph of $$y=\left\lfloor x\right\rfloor$$

3. Infinite Discontinuity

We say $$f$$ has an infinite discontinuity at $$x=a$$ if one of the one-sided limits or both of them are plus or minus infinity.

#### Read More on Infinite Discontinuity

For example, $$f(x)=1/x^{2}$$ has an infinite discontinuity at $$x=0$$ (See Figure 6). Note that we can choose a value for $$f(0)$$ but we cannot make it continuous at $$x=0$$ because $$\lim_{x\to0}f(x)=\infty$$ and $$\infty$$ is not a number to assign it to $$f(0)$$.

 Figure 6: Graph of $$f(x)=\dfrac{1}{x^{2}}$$.

4. Oscillating Discontinuity

#### Read More on Oscillating Discontinuity

For example, $$f(x)=\sin\left(\frac{1}{x}\right)$$ oscillates between $$-1$$ and $$1$$ infinitely often as $$x\to0$$ (See the following figure). Because this function does not approach a single number, it does not have a limit as $$x\to0$$.

 Figure 7: Graph of $$f(x)=\sin\left(\dfrac{1}{x}\right)$$.

#### A function that is discontinuous at every point of its domain

Consider the Dirichlet function defined as $D(x)=\begin{cases} 1 & \text{if }x\text{ is rational}\\ 0 & \text{if }x\text{ is irrational} \end{cases}$ This function is discontinuous at every point; $$D(x)$$ fails to have a limit at any point.

Suppose $$D(x)$$ has a limit $$L$$ at a point $$a$$. Suppose $$\epsilon=1/2$$ is given, we should be able to find a $$\delta>0$$ such that $0<|x-a|<\delta\implies|D(x)-L|<\frac{1}{2}$ Because each deleted neighborhood $$0<|x-a|<\delta$$ contains a rational point $$x_{1}$$ and irrational point $$x_{2}$$, we should have $|D(x_{1})-L|=|1-L|<\frac{1}{2}$ and $|D(x_{2})-L|=|0-L|<\frac{1}{2}$ and hence ${\small 1=|D(x_{1})-L-(D(x_{2})-L)|\leq|D(x_{1})-L|+|D(x_{2})-L|<\frac{1}{2}+\frac{1}{2}}$ Because this is impossible, $$D(x)$$ cannot have a limit.

### Elementary Continuous Functions

Here are some elementary continuous functions

1. Power functions. $$y=x^{n}$$ where $$n$$ is a positive integer is continuous everywhere.

2. Polynomials are continuous everywhere, because for any $$a$$, $$\lim_{x\to a}P(x)=P(a)$$

3. Rational functions. Let $$P(x)$$and $$Q(x)$$ be two polynomials and $$Q(a)\neq0$$. Then the rational function $R(x)=\frac{P(x)}{Q(x)}$ is continuous at $$x=a$$. In other words, a rational function is continuous on their domains and is discontinuous at the points where the denominator is zero.

4. Root functions (1). $$y=\sqrt[n]{x}$$ where $$n$$ is a positive odd integer is continuous everywhere.

5. Root functions (2). $$y=\sqrt[n]{x}$$ where $$n$$ is a positive even integer is continuous on its domain $$[0,\infty)$$.

6. The sine and cosine functions are continuous on $$\mathbb{R}=(-\infty,\infty)$$.
7. The tangent, cotangent, secant and cosecant functions are continuous where they are defined; that is, on their domains. Specifically $$y=\tan x$$ is continuous everywhere except where $$\cos x=0$$; that is when $$x=\frac{\pi}{2}+k\pi$$ for all integers $$k$$. Thus $$y=\tan x$$ is continuous on $\left\{ x\Big|\ x\neq\frac{\pi}{2}+k\pi,\quad k\in\mathbb{Z}\right\}$

 Figure 9: $$y=\tan x$$ is continuous on its domain

The graphs of the rest of trigonometric functions are illustrated in Figure . The domains of these functions are apparent from their graphs.

 (a) $$y=\cot x$$ (b) $$y=\sec x$$ (c) $$y=\text{csc}x$$

Figure 10

8. Inverse trigonometric functions are continuous on their domains. For example, $$y=\arcsin x$$ and $$y=\arccos x$$ are continuous on $$[-1,1]$$and $$y=\arctan x$$ is continuous on $$\mathbb{R}=(-\infty,\infty)$$.

9. Exponential functions. $$y=e^{x}$$ is continuous on its domain $$\mathbb{R}=(-\infty,\infty)$$.

10. Logarithmic functions. $$y=\ln x$$ is continuous on its domain $$(0,\infty)$$.

In fact, most of the functions that we deal with in this course are continuous on their domains. Some exceptions are:

1. The greatest integer function (also known as the floor function) $$f(x)=\left\lfloor x\right\rfloor$$ (or $y=[\![x]\!]$) which is discontinuous at every integer (although it is defined everywhere).

Graph of $$y=\left\lfloor x\right\rfloor$$

2. Piece-wise defined functions are not necessarily continuous on their domains. For example, $y=|x|=\begin{cases} x & \text{if }x\ge0\\ -x & \text{if }x<0 \end{cases}$ is continuous everywhere but the sign function $$y=\text{sgn}(x)$$ and the Heaviside step function $$y=H(x)$$ are continuous everywhere except when $$x=0$$. $\text{sgn}(x)=\begin{cases} 1 & \text{if }x>0\\ 0 & \text{if }x=0\\ -1 & \text{if }x<0 \end{cases}$ $H(x)=\begin{cases} 1 & \text{if }x\geq0\\ 0 & \text{if }x<0 \end{cases}$ [The Heaviside step function is right-continuous at $$x=0$$]

 (a) $$y=|x|$$ is continuous everywhere (b) $$y=\text{sgn}(x)$$ is discontinuous at $$x=0$$ (c) $$y=H(x)$$ is discontinuous at $$x=0$$

Figure 11

Example 2

Let $f(x)=\begin{cases} x^{2}+3 & \text{if }x\geq3\\ mx+5 & \text{if }x<3 \end{cases}$

Solution

For what value of the constant $$m$$ is the function$$f$$ continuous at $$x=3$$? The function$$f$$ is right-continuous at $$x=3$$: \begin{aligned} \lim_{x\to3^{+}}f(x) & =\lim_{x\to3^{+}}(x^{2}+3)\\ & =3^{2}+3=12\\ & =f(3).\end{aligned} Therefore, $$f$$ must be left-continuous at $$x=3$$ to be continuous at $$x=3$$. That is, we must have $$\lim_{x\to3^{-}}f(x)=f(3)$$ or $\lim_{x\to3^{-}}(mx+5)=3m+5=12$ or $m=\frac{7}{3}.$

Example 3

For what values of $$x$$ is there a discontinuity in the graph of $f(x)=\frac{x^{2}-4}{x^{2}-3x+2}?$

Solution

$$f$$ is a rational function, so it is discontinuous at the points where the denominator $$x^{2}-3x+2$$ is zero: $x^{2}-3x+2=0\Rightarrow x=\frac{3\pm\sqrt{3^{2}-4(2)}}{2(1)}$ [Recall that if $$ax^{2}+bx+c=0$$, then $$x=\left(-b\pm\sqrt{b^{2}-4ac}\right)/(2a)$$] $\Rightarrow x=2\ \text{or}\ x=1$ Therefore, $$f$$ is discontinuous at $$x=1$$ and $$x=2$$. However, $$f$$ has a removable discontinuity at $$x=2$$, because $\frac{x^{2}-4}{x^{2}-3x+2}=\frac{(x-2)(x+2)}{(x-1)(x-2)}=\frac{x+2}{x-1}\quad(\text{if }x\neq2).$ [Recall that $$A^{2}-B^{2}=(A-B)(A+B)$$], and $\lim_{x\to2}f(x)=\lim_{x\to2}\frac{x+2}{x-1}=4.$ That is, if we define a new function $g(x)=\begin{cases} \frac{x^{2}-4}{x^{2}-3x+2} & \text{if }x\neq2\\ 4 & \text{if }x=2 \end{cases}$ then $$g$$ is continuous at $$x=2$$, and its graph does not have a hole when $$x=2$$ (compare with the graph of $$f$$ shown in the following figure.

 Figure 12: Graph of $$f(x)=\frac{x^{2}-4}{x^{2}-3x+2}$$.

### Algebraic Operations on Continuous Functions

Theorem 1: If the functions $$f$$ and $$g$$ are continuous at $$x=a$$, and $$k$$ is a number then the following functions are continuous at $$x=a$$:
1. Sum $f(x)+g(x)$ 2. Difference $f(x)-g(x)$ 3. Constant Multiple $kf(x)$ 4. Product $f(x)g(x)$ 5. Quotient $\frac{f(x)}{g(x)}\qquad\text{provided }g(a)\neq0$ 6. Power $\left(f(x)\right)^{n}\qquad n\text{ is a positive integer}$

The results of the above theorem follow from the corresponding Limit Laws. For instance, to prove the sum property we have \begin{aligned} \lim_{x\to a}(f(x)+g(x)) & =\lim_{x\to a}f(x)+\lim_{x\to a}g(x)\\ & =f(a)+g(a)\end{aligned}

### Continuity of Composite Functions

Theorem 2: If $$\lim_{x\to a}g(x)=b$$ and $$f$$ is continuous at the point $$b$$, then \begin{aligned} \lim_{x\to a}f(g(x)) & =f(\lim_{x\to a}g(x))\\ & =f(b).\end{aligned}

Intuitively, the above theorem is plausible because $$\lim_{x\to a}g(x)=b$$ means that when $$x$$ is close to $$a$$, $$g(x)$$ is close to $$b$$ and the continuity of $$f$$at the point $$b$$ means that if the input (which is here $$g(x)$$) is close to $$b$$, the output (which is here $$f(g(x))$$) is close to $$f(b)$$.

#### Show the rigorous proof

The mathematically rigorous proof is as follows. Let $$\epsilon>0$$ be given. We need to find a number $$\delta>0$$ such that for all $$x$$ $0<|x-a|<\delta\Rightarrow\left|f(g(x))-f(b)\right|<\epsilon.$ Because $$f$$ is continuous at $$b$$, we have $\lim_{y\to b}f(y)=f(b)$ and therefore, there exists a number $$\delta_{1}>0$$ such that for all $$y$$ $|y-b|<\delta_{1}\Rightarrow|f(y)-f(b)|<\epsilon$ [Because $$f$$ is continuous, we did not exclude $$y=b$$ and wrote $$|y-b|<\delta_{1}$$ instead of $$0<|y-b|<\delta_{1}$$].
Because $$\lim_{x\to a}g(x)=b$$, there exists a $$\delta>0$$ such that for all $$x$$ $0<|x-a|<\delta\Rightarrow|g(x)-b|<\delta_{1}$ If we let $$y=g(x)$$, we can combine these two statements: For all $$x$$ if $$0<|x-a|<\delta$$ then $$|g(x)-b|<\delta_{1}$$ which implies that $$\left|f(g(x))-f(b)\right|<\epsilon$$. From the definition of limit, it follows that $\lim_{x\to a}f(g(x))=f(b).$

• The theorem holds when $$a$$ is an endpoint of the domain, provided we use an appropriate one-sided limit in place of a two-sided one.

#### Proof of the Root Rule from Sec. 4.4

We can use Theorem 2 to prove Theorem 5 from Section 4.4. Let assume $$\lim_{x\to a}g(x)$$ and the indicated $$n$$th roots exist. Let $$f(x)=\sqrt[n]{x}$$. In Section 4.4, we discussed that (see Theorem 4) $\lim_{x\to a}f(x)=\lim_{x\to a}\sqrt[n]{x}=\sqrt[n]{a}$ That is, $$f(x)$$ is continuous on its domain. Applying Theorem 2 which states that $\lim_{x\to a}f(g(x))=f\left(\lim_{x\to a}g(x)\right)$ we obtain $\lim_{x\to a}\sqrt[n]{g(x)}=\sqrt[n]{\lim_{x\to a}g(x)}.$

Example 4

Evaluate the following limits
(a) $${\displaystyle \lim_{x\to-3}\sin\left(\frac{x+3}{x^{2}+3x}\right)}$$
(b) $${\displaystyle \lim_{x\to1}\arctan\left(x+\cos\left(\frac{\pi}{2}x\right)\right)}$$
(c) $${\displaystyle \lim_{x\to0}\ln\left(1+e^{\tan x}\right)}$$

Solution

(a) Because $$y=\sin x$$ is a continuous function everywhere, by Theorem 2 we have $\lim_{x\to-3}\sin\left(\frac{x+3}{x^{2}+3x}\right)=\sin\left(\lim_{x\to-3}\frac{x+3}{x^{2}+3x}\right)$ If we substitute $$x=-3$$ in $$(x+3)/(x^{2}+3x)$$ we get $$0/0$$, which shows the numerator and the denominator have a common factor: $\frac{x+3}{x^{2}+3x}=\frac{x+3}{x(x+3)}=\frac{1}{x}\qquad(x\neq-3)$ Therefore, \begin{aligned} \sin\left(\lim_{x\to-3}\frac{x+3}{x^{2}+3x}\right) & =\sin\left(\lim_{x\to-3}\frac{\cancel{x+3}}{x\cancel{(x+3)}}\right)\\ & =\sin\left(\lim_{x\to-3}\frac{1}{x}\right)\\ & =\sin\left(\frac{1}{-3}\right)\\ & =\sin\left(\frac{-1}{3}\right)\approx-0.3272.\end{aligned}

(b) Because $$y=\arctan x$$ is a continuous function, we have \begin{aligned} \lim_{x\to1}\arctan\left(x+\cos\left(\frac{\pi}{2}x\right)\right) & =\arctan\left(\lim_{x\to1}\left(x+\cos\left(\frac{\pi}{2}x\right)\right)\right)\\ & =\arctan\left(\lim_{x\to1}x+\lim_{x\to1}\cos\left(\frac{\pi}{2}x\right)\right)\end{aligned} and again because $$y=\cos x$$ is a continuous function, we have \begin{aligned} \arctan\left(\lim_{x\to1}x+\lim_{x\to1}\cos\left(\frac{\pi}{2}x\right)\right) & =\arctan\left(\lim_{x\to1}x+\cos\left(\lim_{x\to1}\frac{\pi}{2}x\right)\right)\\ & =\arctan\left(1+\cos\left(\frac{\pi}{2}\right)\right)\\ & =\arctan\left(1+0\right)\\ & =\frac{\pi}{4}.\end{aligned} Therefore, $\lim_{x\to1}\arctan\left(x+\cos\left(\frac{\pi}{2}x\right)\right)=\frac{\pi}{4}.$

(c) Because $$y=\ln x$$ is a continuous function on its domain $$(0,\infty)$$, we have \begin{aligned} \lim_{x\to0}\ln\left(1+e^{\tan x}\right) & =\ln\left(\lim_{x\to0}\left(1+e^{\tan x}\right)\right)\\ & =\ln\left(\lim_{x\to0}1+\lim_{x\to0}e^{\tan x}\right)\end{aligned} Because $$y=e^{x}$$ is continuous everywhere, \begin{aligned} \ln\left(\lim_{x\to0}1+\lim_{x\to0}e^{\tan x}\right) & =\ln\left(\lim_{x\to0}1+e^{\lim_{x\to0}\tan x}\right)\\ & =\ln(1+e^{0})\\ & =\ln(1+1)=\ln2.\end{aligned} That is, $\lim_{x\to0}\ln\left(1+e^{\tan x}\right)=\ln2.$

Example 5

Find $\lim_{x\to\infty}\sin\frac{1}{x}.$

Solution

Because $$f(x)=\sin x$$ is continuous everywhere, $\lim_{x\to\infty}\sin\frac{1}{x}=\sin\left(\lim_{x\to\infty}\frac{1}{x}\right).$ [Recall that $$\lim_{x\to\pm\infty}\frac{1}{x^{r}}=0$$ where $$r>0$$ is a rational number (Theorem 1 in Section 4.7)]. Thus $\sin\left(\lim_{x\to\infty}\frac{1}{x}\right)=\sin(0)=0.$ The graph of $$y=\sin\frac{1}{x}$$ is shown below.

 Figure 13: Graph of $$y=\sin\frac{1}{x}$$. We can see from this figure that $$\lim_{x\to\pm\infty}\sin\frac{1}{x}=0$$.

Example 6

Find
(a) $${\displaystyle \lim_{x\to0^{+}}\arctan\left(\frac{1}{x}\right)}$$
(b) $${\displaystyle \lim_{x\to0^{-}}\arctan\left(\frac{1}{x}\right)}$$
(c) $${\displaystyle \lim_{x\to\infty}\arctan\left(\frac{1}{x}\right)}$$
(d) $${\displaystyle \lim_{x\to-\infty}\arctan\left(\frac{1}{x}\right)}$$

Solution

(a) Because $$y=\arctan x$$ is a continuous function, we can find the limit of its input first. That is, \begin{aligned} \lim_{x\to0^{+}}\arctan\left(\frac{1}{x}\right) & =\arctan\left(\lim_{x\to0^{+}}\frac{1}{x}\right)\\ & =\arctan(+\infty)\\ & =\frac{\pi}{2}\end{aligned} [To be accurate, we cannot write $$\arctan(+\infty)$$ because $$+\infty$$ is not a number] (b) Similar to part (a) \begin{aligned} \lim_{x\to0^{-}}\arctan\left(\frac{1}{x}\right) & =\arctan\left(\lim_{x\to0^{-}}\frac{1}{x}\right)\\ & =\arctan(-\infty)\\ & =-\frac{\pi}{2}.\end{aligned}

(c) \begin{aligned} \lim_{x\to+\infty}\arctan\left(\frac{1}{x}\right) & =\arctan\left(\lim_{x\to+\infty}\frac{1}{x}\right)\\ & =\arctan(0)\\ & =0.\end{aligned}

(d) \begin{aligned} \lim_{x\to-\infty}\arctan\left(\frac{1}{x}\right) & =\arctan\left(\lim_{x\to-\infty}\frac{1}{x}\right)\\ & =\arctan(0)\\ & =0.\end{aligned} The graph of $$y=\arctan(1/x)$$ is shown in the following figure.

 Figure 14: Graph of $$y=\arctan\frac{1}{x}$$.

It follows from Theorem 2 that

Theorem 3: If $$g$$ is a continuous function at $$a$$ and if $$f$$is continuous at $$g(a),$$then $$f\circ g$$ is continuous at $$a$$

#### Show the proof

To prove the above theorem, we need to show $\lim_{x\to a}(f\circ g)(x)=(f\circ g)(a)$ or $\lim_{x\to a}f(g(x))=f(g(a)).$ Because $$g$$ is continuous at $$x=a$$, then $\lim_{x\to a}g(x)=g(a)$ and because $$f$$ is continuous at $$g(a),$$ by \begin{aligned} \lim_{x\to a}f(g(x)) & =f\left(\lim_{x\to a}g(x)\right)\\ & =f(g(a)).\end{aligned}

Example 7
Where are the following functions continuous?
(a) $$h(x)=\cos(x^{3}+1)$$
(b) $$F(x)=\left|\dfrac{x\cos x}{x^{2}+1}\right|$$
(c) $$G(x)=\sqrt{x^{2}+2x-2}$$

Solution

(a) We have $$h(x)=f(g(x))$$ where $f(x)=\cos x\ \text{and}\ g(x)=x^{3}+1.$ Because both $$f$$ and $$g$$ (which is a polynomial) are continuous on $$\mathbb{R}$$, it follows from Theorem 3 that $$h=f\circ g$$ is also continuous on $$\mathbb{R}$$.

(b) Notice that $$F$$ is the composition of the following functions $$F=f\circ g$$: $f(x)=|x|\ \text{and}\ g(x)=\frac{x\cos x}{x^{2}+1}$ Obviously $$f$$ is continuous on $$\mathbb{R}$$. We can write $$g(x)=u(x)v(x)/w(x)$$ where $u(x)=x,\ v(x)=\cos x,\ w(x)=x^{2}+1$ Because $$u,v$$ and $$w$$ are continuous functions on $$\mathbb{R}$$ and $$w(x)\neq0$$, by the product and quotient rules (Theorem 1), $$g$$ is continuous on $$\mathbb{R}$$ too. Because $$f$$ and $$g$$ are both continuous on $$\mathbb{R}$$ and $$F=f\circ g$$, it follows from Theorem 3 that $$F$$ is also continuous on $$\mathbb{R}$$.

(c) We have $$G=f\circ g$$ where $f(x)=\sqrt{x},\ \text{and}\ g(x)=x^{2}+2x-2$ The square root function is continuous on its domain $$[0,\infty)$$, and $$g$$ which is a polynomial is continuous on $$\mathbb{R}$$. So by Theorem 3, $$G$$ is continuous on its domain, which is $\left\{ \left.x\in\mathbb{R}\right|\ x^{2}+2x-2\geq0\right\}$ We can use the sign table to determine where $$g(x)=x^{2}+2x-2\geq0$$: $x^{2}+2x-2=0\Rightarrow x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-2\pm\sqrt{2^{2}+4\times2}}{2}=-1\pm\sqrt{3}$ So we can write $x^{2}+2x-2=(x+1-\sqrt{3})(x+1+\sqrt{3})$

$\large&space;x$ $$-\infty$$   $$-1-\sqrt{3}$$   $$-1+\sqrt{3}$$   $$\infty$$
sign of $$(x+1-\sqrt{3})$$   – – – – – – – – $0$ + + + +
sign of $$(x+1+\sqrt{3})$$   – – – – $0$ + + + + + + + + +
$$\therefore$$ sign of $$x^{2}+2x-2$$   + + + + $0$ – – – – $0$ + + + +

Therefore, $G$ is continuous on the following set: $\left\{ \left.x\in\mathbb{R}\right|\ x^{2}+2x-2\geq0\right\} =(-\infty,-1-\sqrt{3}]\cup[-1+\sqrt{3},\infty)$