Continuity

Continuity, most of the functions we study in elementary calculus are described by simple formulas. These functions almost always possess derivatives and, in fact, a portion of any first course in calculus is devoted to the development of routine methods for computing derivatives. However, not all functions possess derivatives everywhere. For example, the functions $\frac{1+x^2}{x},\ \cot x\ and\ \sin \left(\frac{1}{x}\right)$ do not possess derivatives at x = 0 no matter how they are defined at x = 0.
As we progress in the study of analysis, it is important to enlarge substantially the class of functions under examination. Functions which possess derivatives everywhere form a rather restricted class; extending this class to functions which are differentiable except at a few isolated points does not enlarge it greatly. We wish to investigate significantly larger classes of functions, and to do so we introduce the notion of a continuous functions.

Definitions

Suppose that ƒ is a function from a domain D in $R^1$ to $R^1$ . The function ƒ is continuous at a if and only if (i) the point a is in an open interval I contained in D, and (ii) for each positive number ε & there is a positive number δ such that

$\left|f\left(x\right)-f\left(a\right)\right|<\epsilon \ \ \ \ \ \ \ \ \ \ \ \ whenever\ \ \ \ \ \left|x-a\right|<\delta$

If f is continuous at each point of a set S, we say that ƒ is continuous on S. A function ƒ is called continuous if it is continuous at every point of its domain. The geometric significance of continuity at a point a is indicated in Figure 2.1. We recall that the inequality $\left|f\left(x\right)-f\left(a\right)\right|<\epsilon$ is equivalent to the double inequality

$-\epsilon <f\left(x\right)-f\left(a\right)<\epsilon$

or,

$f\left(a\right)-\epsilon <f\left(x\right)<f\left(a\right)+\epsilon$

Similarly, the inequality $\left|x-a\right|<\delta$ is equivalent to the two inequalities

$a-\delta <x<a+\delta$

We construct the four lines $\ x=a-\delta ,\ x=a+\delta ,\ y=f\left(a\right)-\epsilon ,\ y=f\left(a\right)+\epsilon$ as shown in Figure 2.1. The rectangle determined by these four lines has its center at the point with coordinates (a, f(a)). The geometric interpretation of continuity at a point may be given in terms of this rectangle. A function ƒ is continuous at a if for each $\\epsilon >0$ there is a number $\ \delta >0$ such that the graph of ƒ remains within the rectangle for all x in the interval $\left(a-\delta ,\ a+\delta \right)$ .
It is usually very difficult to verify continuity directly from the definition. Such verification requires that for every positive number $\epsilon$ , we exhibit a number $\delta$ and show that the graph of ƒ lies in the appropriate rectangle. However, if the function f is given by a sufficiently simple expression, it is sometimes possible to obtain an explicit value for the quantity $\delta$ corresponding to a given number $\epsilon$ . We describe the method by means of two examples.

Example 1.

Given the function

$f:\ x\ \longrightarrow \ \frac{1}{x+1},\ \ \ x\ \ne -1$ ,

and $a=1,\epsilon =0.1$ , find a number $\delta$ such that

$\left|f\left(x\right)-f\left(1\right)\right|<0.1$ for $\ \left|x-1\right|<\delta$ .

Figure 2.2

Solution

We sketch the graph of ƒ and observe that ƒ is decreasing for x > -1 (see Figure 2.2). The equations f(x) − f(1) = 0.1, f(x) – f(1) = -0.1 can be solved for x. We find

$\begin{array}{l}\frac{1}{x+1}-\frac{1}{2}=0.1\ \ \longleftrightarrow \ \ x=\frac{2}{3}\\ \frac{1}{x+1}-\frac{1}{2}=-0.1\ \longleftrightarrow \ x=\frac{3}{2}\end{array}$

Since ƒ is decreasing in the interval $\frac{2}{3}<x<\frac{3}{2}$ , it is clear that the graph of ƒ lies in the rectangle formed by the lines $\ x=\frac{2}{3},\ x=\frac{3}{2},\ y=\frac{1}{2}-0.1,\ y=\frac{1}{2}+0.1$ . Since the distance from x = 1 to x = 2/3 is smaller than the distance from x = 1 to x = 3/2, we select $\delta =1-\frac{2}{3}=\frac{1}{3}$ . We make the important general observation that when a value of $\delta$ is obtained for a given quantity ε, then any smaller (positive) value for $\delta$ may also be used for the same number ε.

Remarks.

For purposes of illustration, we assume in this chapter that $\sqrt[n]{x}$ is defined for all $\ x\ge 0$ if n is even and that $\sqrt[n]{x}$ is defined for all x if n is odd. We also suppose that for positive numbers $\ x_1,\ x_2$ , the inequality $\sqrt{x_2}>\sqrt{x_1}$ holds whenever $\ x_2>x_1$ . Facts of this type concerning functions of the form $\sqrt[n]{x}$ , with n a natural number, do not follow from the results given thus far.

Figure 2.3

Example 2.

Consider the function

$f:\ x\ \longrightarrow \ \left\{\begin{matrix}\frac{x-4}{\sqrt{x}-2}&\ \ \ x\ge 0,\ x\ \ne 4,\\ 4&\ \ \ \ x=4.\end{matrix}\right\}$
If $\epsilon$ = 0.01, find a $\delta$ such that $\left|f\left(x\right)-f\left(4\right)\right|<0.01$ for all x such that $\left|x-4\right|<\delta$ .

Solution

If $x\ \ne 4$ , then

$f\left(x\right)=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-2}=\sqrt{x}+2$

The graph of ƒ is shown in Figure 2.3, and we observe that ƒ is an increasing function. We solve the equations f(x) – f(4) = 0.01 and f(x) − f(4) = -0.01 and obtain

$\sqrt{x}+2-4=0.01\ \longleftrightarrow \ \ \sqrt{x}=2.01\ \ \longleftrightarrow \ \ x=4.0401$

$\sqrt{x}+2-4=-0.01\ \longleftrightarrow \ \ \sqrt{x}=1.99\ \ \longleftrightarrow \ \ x=3.9601$

Since ƒ is increasing, it follows that $\left|f\left(x\right)-f\left(4\right)\right|<0.01$ for 3.9601 <x< 4.0401. Selecting $\delta \$ = 0.0399, we find that $\left|f\left(x\right)-f\left(4\right)\right|<\epsilon \$ for $\left|x-4\right|<\delta$ .
We shall frequently be concerned with functions which have a domain in $R^1$ which consists of an interval except for a single point. This exceptional point may be an interior point or an endpoint of the interval. For example, the function $\frac{\ x^2+1\ }{x}$ is defined for all x except x = 0. The function log x is defined for x>0 but not for x = 0. Other examples are cot x, defined on any interval not containing an integral multiple of $\pi$ , and the function $\frac{\ x^3+8\ }{x+2}$ not defined for x = -2; in fact, any function defined as the quotient of two polynomials has excluded from its domain those values at which the denominator is zero.

Definition

Suppose that a and L are real numbers and ƒ is a function from a domain D in $R^1$ to $R^1$ . The number a may or may not be in the domain of f. The function f tends to L as a limit as x tends to a if and only if (i) there is an open interval I containing a which, except possibly for the point a, is contained in D, and (ii) for each positive number $\epsilon$ there is a positive number $\delta$ such that

$\left|f\left(x\right)-L\right|<\epsilon$ whenever $\ 0<\left|x-a\right|<\delta \$

If ƒ tends to L as x tends to a, we write

$f\left(x\ \right)\longrightarrow L$ as $x\ \longrightarrow a$

and denote the number L by

$\ \ \lim _{x\to a}f\left(x\right)$

or by

$\ \ \lim _{x\to a}f\left(x\right)$

Remarks

(i) We see that a function f is continuous at a if and only if a is in the domain of ƒ and $f\left(x\right)\longrightarrow f\left(a\right)$ as $\ x\longrightarrow a$ .
(ii) The condition $0<\left|x-a\right|<\delta$ (excluding the possibility x = a) is used rather than the condition $\left|x-a\right|<\delta$ as in the definition of continuity since f may not be defined at a itself.