Table of Contents

## Chapter 4

Limit and continuity

One of the key concepts of calculus is the limit concept. Calculus has two major subfields: differential calculus and integral calculus. Differential calculus is concerned about the rate of change and finding tangent lines to curves. Integral calculus is about computing the total area under a curve and measuring the total effect of a process of continuous change. These concepts are defined in terms of limits. In fact, every single notion of calculus is a limit in one way or another, and what distinguishes calculus from algebra is the concept of limit.

The concept of limit was implicitly used by the ancient Greek scholars. However, it was Isaac Newton in the 17th century who explicitly talked about limits. In the 18th century, the French mathematician, Jean le Rond d’Alembert (1717–1783), used the notion of limits and tried to provide a definition for it. Finally, another French mathematician Augustin-Louis Cauchy (1789–1857) formulated a satisfactory mathematical definition of a limit, which is still used today.

In this chapter, we introduce the concepts of limit and continuity at first simply and intuitively,
and then with more careful argument. We will learn how to evaluate many limits, but one of the
major techniques of finding limits will be discussed when we talk about applications of
differentiation.

• In the 1960s an alternative formulation of calculus, called non-standard analysis, was introduced. Non-standard analysis legitimates the concept of infinitesimals, which was vaguely used by Gottfried Leibnitz (1646–1716), Leonhard Euler (1707–1783), and many others in the beginnings of calculus. For more information read the Wikipedia article on non-standard analysis.

### 4.1 The concept of a limit

### 4.2 The precise definition of a limit

### 4.3 One-sided limits

### 4.4 Section 4.4

### 4.5 Section 4.5

### 4.6 Section 4.6

### 4.7 Section 4.7

### 4.8 Continuity

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

Here is the formal definition of continuity.

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 \(a\).

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

- 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 \(a\).

- 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 \(b\).

Solution 1. For \(-2<a<2\), we have \begin{align*} \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}}\tag{by Difference Rule in Sec. 4.4}\\ & =3-\sqrt{\lim _{x\to a}(4-x^{2})}\tag{by Root Rule}\\ & =3-\sqrt{4-a^{2}}\tag{\ensuremath{(4-x^{2})}is a Polynomial}\\ & =f(a) \end{align*}

Similarly, we can show \[ \lim _{x\to -2^{+}}f(x)=3=f(-2),\quad \lim _{x\to 2^{-}}f(x)=3=f(2) \] 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 \(f(a).\)

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 \[ \lim _{x\to 2}(x+2)=4, \] we obtain \[ \lim _{x\to 2}\frac{x^{2}-4}{x-2}=4. \] 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 4.8.3).

(a) | (b) |

##### 2. Jump discontinuity

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

For example, the Heaviside step function \(H(x)\) is defined as \[ H(x)=\begin{cases} 1 & \text{if }x\geq 0\\ 0 & \text{if }x<0 \end{cases} \] This function has a jump discontinuity at \(x=0\) because \[ \lim _{x\to 0^{-}}H(x)=0\neq \lim _{x\to 0^{+}}H(x)=1. \] We note that \(H(x)\) is right continuous at \(x=0\) because \(H(0)=\lim _{x\to 0^{+}}H(x)=1\). (See Figure 4.8.4).

The greatest integer function (also known as the floor function) \(y=\left \lfloor x\right \rfloor \) (or \(y=\left \llbracket x\right \rrbracket \)) has a jump discontinuity at every integer (see Figure 4.8.5). For example, \[ \lim _{x\to 2^{-}}\left \lfloor x\right \rfloor =1\neq \lim _{x\to 2^{+}}\left \lfloor x\right \rfloor =2. \]

##### 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.

For example, \(f(x)=1/x^{2}\) has an infinite discontinuity at \(x=0\) (See Figure 4.8.6). Note that we can choose a value for \(f(0)\) but we cannot make it continuous at \(x=0\) because \(\lim _{x\to 0}f(x)=\infty \) and \(\infty \) is not a number to assign it to \(f(0)\).

##### 4. Oscillating discontinuity

For example, \(f(x)=\sin \left (\frac{1}{x}\right )\) oscillates between \(-1\) and \(1\) infinitely often as \(x\to 0\) (See Figure 4.8.7). Because this function does not approach a single number, it does not have a limit as \(x\to 0\).

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 \[ 1=|D(x_{1})-L-(D(x_{2})-L)|\leq |D(x_{1})-L|+|D(x_{2})-L|<\frac{1}{2}+\frac{1}{2} \] Because this is impossible, \(D(x)\) cannot have a limit.

#### Elementary continuous functions

Here are some elementary continuous functions

- 1.
- \(y=x^{n}\) 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.
- Let \(P(x)\)and \(Q(x)\) be two polynomials and \(Q(a)\neq 0\). 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.
- \(y=\sqrt [n]{x}\) where \(n\) is a positive odd integer is continuous everywhere.
- 5.
- \(y=\sqrt [n]{x}\) 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.
- 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 \[ \left \{ x\Big |\ x\neq \frac{\pi }{2}+k\pi ,\quad k\in \mathbb{Z}\right \} \]
The graphs of the rest of trigonometric functions are illustrated in Figure . The domains of these functions are apparent from their graphs.

- 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.
- \(y=e^{x}\) is continuous on its domain \(\mathbb{R}=(-\infty ,\infty )\).
- 10.
- \(y=\ln x\) 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) \(f(x)=\left \lfloor x\right \rfloor \) which is discontinuous at every integer (although it is defined everywhere).
- 2.
- Piece-wise defined functions are not necessarily continuous on their domains. For example, \[ y=|x|=\begin{cases} x & \text{if }x\ge 0\\ -x & \text{if }x<0 \end{cases} \] is continuous everywhere but the sign function \(y=\text{sgn}(x)\) and the Heaviside step function \(y=H(x)\) are continuous everywhere except when \(x=0\). \[ \text{sgn}(x)=\begin{cases} 1 & \text{if }x>0\\ 0 & \text{if }x=0\\ -1 & \text{if }x<0 \end{cases} \] \[ H(x)=\begin{cases} 1 & \text{if }x\geq 0\\ 0 & \text{if }x<0 \end{cases} \] [The Heaviside step function is right-continuous at \(x=0\)]

Example 2. Let \[ f(x)=\begin{cases} x^{2}+3 & \text{if }x\geq 3\\ mx+5 & \text{if }x<3 \end{cases} \] For what value of the constant \(m\) is the function\(f\) continuous at \(x=3\)?

Solution 2. The function\(f\) is right-continuous at \(x=3\): \begin{align*} \lim _{x\to 3^{+}}f(x) & =\lim _{x\to 3^{+}}(x^{2}+3)\\ & =3^{2}+3=12\\ & =f(3). \end{align*}

Therefore, \(f\) must be left-continuous at \(x=3\) to be continuous at \(x=3\). That is, we must have \(\lim _{x\to 3^{-}}f(x)=f(3)\) or \[ \lim _{x\to 3^{-}}(mx+5)=3m+5=12 \] 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 3. \(f\) 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 \[ \frac{x^{2}-4}{x^{2}-3x+2}=\frac{(x-2)(x+2)}{(x-1)(x-2)}=\frac{x+2}{x-1}\quad (\text{if }x\neq 2). \] [Recall that \(A^{2}-B^{2}=(A-B)(A+B)\)], and \[ \lim _{x\to 2}f(x)=\lim _{x\to 2}\frac{x+2}{x-1}=4. \] That is, if we define a new function \[ g(x)=\begin{cases} \frac{x^{2}-4}{x^{2}-3x+2} & \text{if }x\neq 2\\ 4 & \text{if }x=2 \end{cases} \] 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 Figure 4.8.12).

#### 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 \[ \frac{f(x)}{g(x)}\qquad \text{provided }g(a)\neq 0 \] 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 in Theorem ??. For instance, to prove the sum property we have \begin{align*} \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{align*}

#### Continuity of composite functions

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 \(f(b)\).

The mathematically rigorous proof is as follows.

Proof. 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 Theorem ??: We can use Theorem 2 to prove Theorem ?? from Section ??. Let assume \(\lim _{x\to a}g(x)\) and the indicated \(n\)th roots exist. Let \(f(x)=\sqrt [n]{x}\). In Section ??, we discussed that (see Theorem ??) \[ \lim _{x\to a}f(x)=\lim _{x\to a}\sqrt [n]{x}=\sqrt [n]{a} \] 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) \(\displaystyle \lim _{x\to 1}\arctan \left (x+\cos \left (\frac{\pi }{2}x\right )\right )\)

(c) \(\displaystyle \lim _{x\to 0}\ln \left (1+e^{\tan x}\right )\)

Solution 4. (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{align*} \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{align*}

(b) Because \(y=\arctan x\) is a continuous function, we have \begin{align*} \lim _{x\to 1}\arctan \left (x+\cos \left (\frac{\pi }{2}x\right )\right ) & =\arctan \left (\lim _{x\to 1}\left (x+\cos \left (\frac{\pi }{2}x\right )\right )\right )\\ & =\arctan \left (\lim _{x\to 1}x+\lim _{x\to 1}\cos \left (\frac{\pi }{2}x\right )\right ) \end{align*}

and again because \(y=\cos x\) is a continuous function, we have \begin{align*} \arctan \left (\lim _{x\to 1}x+\lim _{x\to 1}\cos \left (\frac{\pi }{2}x\right )\right ) & =\arctan \left (\lim _{x\to 1}x+\cos \left (\lim _{x\to 1}\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{align*}

Therefore, \[ \lim _{x\to 1}\arctan \left (x+\cos \left (\frac{\pi }{2}x\right )\right )=\frac{\pi }{4}. \] (c) Because \(y=\ln x\) is a continuous function on its domain \((0,\infty )\), we have \begin{align*} \lim _{x\to 0}\ln \left (1+e^{\tan x}\right ) & =\ln \left (\lim _{x\to 0}\left (1+e^{\tan x}\right )\right )\\ & =\ln \left (\lim _{x\to 0}1+\lim _{x\to 0}e^{\tan x}\right ) \end{align*}

Because \(y=e^{x}\) is continuous everywhere, \begin{align*} \ln \left (\lim _{x\to 0}1+\lim _{x\to 0}e^{\tan x}\right ) & =\ln \left (\lim _{x\to 0}1+e^{\lim _{x\to 0}\tan x}\right )\\ & =\ln (1+e^{0})\\ & =\ln (1+1)=\ln 2. \end{align*}

That is, \[ \lim _{x\to 0}\ln \left (1+e^{\tan x}\right )=\ln 2. \]

Solution 5. 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 ??)]. Thus \[ \sin \left (\lim _{x\to \infty }\frac{1}{x}\right )=\sin (0)=0. \] The graph of \(y=\sin \frac{1}{x}\) is shown in Figure 4.8.13.

Example 6. Find

(a) \(\displaystyle \lim _{x\to 0^{+}}\arctan \left (\frac{1}{x}\right )\)

(b) \(\displaystyle \lim _{x\to 0^{-}}\arctan \left (\frac{1}{x}\right )\)

(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 6. (a) Because \(y=\arctan x\) is a continuous function, we can find the limit of its input first. That is, \begin{align*} \lim _{x\to 0^{+}}\arctan \left (\frac{1}{x}\right ) & =\arctan \left (\lim _{x\to 0^{+}}\frac{1}{x}\right )\\ & =\arctan (+\infty )\\ & =\frac{\pi }{2} \end{align*}

[To be accurate, we cannot write \(\arctan (+\infty )\) because \(+\infty \) is not a number]

(b) Similar to part (a) \begin{align*} \lim _{x\to 0^{-}}\arctan \left (\frac{1}{x}\right ) & =\arctan \left (\lim _{x\to 0^{-}}\frac{1}{x}\right )\\ & =\arctan (-\infty )\\ & =-\frac{\pi }{2}. \end{align*}

(c) \begin{align*} \lim _{x\to +\infty }\arctan \left (\frac{1}{x}\right ) & =\arctan \left (\lim _{x\to +\infty }\frac{1}{x}\right )\\ & =\arctan (0)\\ & =0. \end{align*}

(d) \begin{align*} \lim _{x\to -\infty }\arctan \left (\frac{1}{x}\right ) & =\arctan \left (\lim _{x\to -\infty }\frac{1}{x}\right )\\ & =\arctan (0)\\ & =0. \end{align*}

The graph of \(y=\arctan (1/x)\) is shown in the following figure.

It follows from Theorem 2 that

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 Theorem 2 \begin{align*} \lim _{x\to a}f(g(x)) & =f\left (\lim _{x\to a}g(x)\right )\\ & =f(g(a)). \end{align*} □

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 7. (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 \(w(x)\neq 0\), 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 \[ \left \{ \left .x\in \mathbb{R}\right |\ x^{2}+2x-2\geq 0\right \} \] We can use the sign table to determine where \(g(x)=x^{2}+2x-2\geq 0\): \[ x^{2}+2x-2=0\Rightarrow x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-2\pm \sqrt{2^{2}+4\times 2}}{2}=-1\pm \sqrt{3} \] So we can write \[ x^{2}+2x-2=(x+1-\sqrt{3})(x+1+\sqrt{3}) \]

\(x\) | \(-\infty \) |
| \(-1-\sqrt{3}\) |
| \(-1+\sqrt{3}\) |
| \(\infty \) |

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 \[ \left \{ \left .x\in \mathbb{R}\right |\ x^{2}+2x-2\geq 0\right \} =(-\infty ,-1-\sqrt{3}]\cup [-1+\sqrt{3},\infty ) \]