Calculus I 01.04 Continuity and One-Sided Limits

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Continuity at a Point and on an Open Interval

• Determine continuity at a point and continuity on an open interval.
• Determine one-sided limits and continuity on a closed interval.
• Use continuity properties.
• Describe the Intermediate Value Theorem.

A function \(f\) is mathematically continuous at \(x=c\) if the graph for \(f\) has no interruption, holes, jumps, or gaps at \(f(c)\). Figure 1.4.1 describes three values for \(x\) at which the graph for \(f\) is not continuous. At all other points on the interval \((a,b)\), the graph is uninterrupted and continuous.

Three conditions exist where the graph for \(f\) is not continuous at \(x=c\).
Figure 1.4.1

Figure 1.4.1 shows that continuity at \(x=c\) can be destroyed by three conditions.

1. The function is not defined at \(x=c\).
2. The limit for \(f(x)\) does not exist at \(x=c\).
3. The limit for \(f(x)\) exists at \(x=c\), but is not equal to \(f(c)\).

If none of the three conditions is true, then the function \(f\) is called continuous at \(\mathbf{c}\), as described in the definition below.

Definition 1.4.1 Continuity Definition

A function \(f\) is continuous at \(\mathbf{c}\) when three conditions are met.

1. \(f(c)\) is defined.

2. \(\lim_{x \to c} f(x)\) exists.

3. \(\lim_{x \to c} f(x)=f(c)\)

Continuity on an Open Interval

A function is continuous on an open interval \(\mathbf{(a,b)}\) when the function continuous at each point on the interval. A function that is continuous on the entire real number line \((-\infty ,\infty )\) is everywhere continuous.

Figure 1.4.2

Consider an open interval \(I\) that contains a real number \(c\). If a function \(f\) is defined on \(I\) (except possibly at \(c\)), and \(f\) is not continuous at \(c\), then \(f\) is said to have a discontinuity at \(c\). Discontinuities fall into two categories: removable and non removable. A discontinuity at \(c\) is called removable when \(f\) can be made continuous by appropriately defining (or redefining) \(f(c)\). For instance, the functions shown in Figures 1.4.2(a) and (c) have removable discontinuities at \(c\) and the function shown in Figure 1.4.2(b) has a non removable discontinuity at \(c\). Example 1.4.1 Function Continuity Define the continuity for each function. $$\mathbf{\text{a.}}\: f(x)=\frac{1}{x}\:\:\:\: \mathbf{ \text{b.}}\:g(x)=\frac{x^2-1}{x-1}\:\:\:\: \mathbf{ \text{c.}}\:h(x)=\left\{\begin{matrix} x+1 &x \leq0 \\ x^2+1 & x >0 \end{matrix}\right. \:\:\:\:\mathbf{ \text{d.}}\:y= \sin x$$ a. The domain for \(f\) is all nonzero real numbers. Applying Theorem 1.3.3 yields that \(f\) is continuous at every \(x\)-value in its domain. At \(x=0\), \(f\) has a non removable discontinuity, as shown in Figure 1.4.3(a). In other words, there is no way to define \(f(0)\) so as to make the function continuous at \(x=0\). b. The domain for \(g\) is all real numbers except for \(x=1\). Applying Theorem 1.3.3 yields that \(g\) is continuous at every \(x\)-value in its domain. At \(x=1\), the function has a removable discontinuity, as shown in Figure 1.4.3(b). By defining \(g(1)\) as 2, the “redefined” function is continuous for all real numbers. c. The domain for \(h\) is all real number. The function \(h\) is continuous on \((-\infty,0)\) and \((\infty,0)\), and, because $$\lim_{x \to 0} h(x)=1$$ \(h\) is continuous on the entire real number line, as shown in Figure 1.4.3(c). d. The domain for \(y\) is all real numbers. Applying Theorem 1.3.6 yields that the function is continuous on its entire domain \((-\infty,\infty)\), as shown in Figure 1.4.3(d) Figure 1.4.3

One-Sided Limits and Continuity on a Closed Interval

Figure 1.4.4

To understand continuity on a closed interval first look at a new limit called one-sided limit. For example, the limit from the right (or right-hand limit) means that as \(x\) approaches \(c\) from values greater than \(c\) [see Figure 1.4.4(a)]. This limit is denoted as $$\lim_{x \to c^+} f(x)=L.$$      Limit from the right Similarly, the limit from the left (or left-hand limit) means that \(x\) approaches \(c\) from values less than \(c\) [see Figure 1.4.4(b)]. This limit is denoted as style="vertical-align:middle;" $$\lim_{x \to c^-} f(x)=L.$$      Limit from the left One-sided limits are useful in taking limits for functions involving radicals. For example, if \(n\) is an even integer, then $$\lim_{x \to 0^+} \sqrt[n]{x}=0.$$

Example 1.4.2 A One-Sided Limit from the Right

The limit for \(f(x)\) as \(x\) approaches -2 from the right is 0. Figure 1.4.5

Find the limit for \(f(x)=\sqrt{4-x^2}\) as \(x\) approaches -2 from the right. Solution As shown in Figure 1.4.5, the limit as \(x\) approaches -2 from the right is $$\lim_{x \to -2^+} \sqrt{4-x^2}=0.$$ One-sided limits can be used to investigate behavior in step functions. One common step function is the greatest integer function [\![x]\!], defined as $$[\![x]\!]=\:\:\text{greatest integer} \: n \: \text{such that}\:n \leq x.$$     Greatest integer function For example, \([\![2.5]\!]=2\) and \([\![-2.5]\!]=-3\).

Example 1.4.3 The Greater Integer Function

Greater integer function Figure 1.4.6

Find the limit for the greatest integer function \(f(x)=[\![x]\!]\) as \(x\) approaches 0 from the left and from the right. Solution As shown in Figure 1.4.6, the limit as \(x\) approaches 0 from the left is $$\lim_{x \to 0^-} [\![x]\!]=-1$$ and the limit as \(x\) approaches 0 from the right is $$\lim_{x \to 0^+} [\![x]\!]=0.$$ The greatest integer function has a discontinuity at zero because the left- and right-handed limits at zero are different. By similar reasoning, the greater integer function has a discontinuity at any integer \(n\).

Continuous function on a closed interval. Figure 1.4.7

Theorem 1.4.1 Evidence a Limit Exists Let \(f\) be a function, and let \(c\) and \(L\) be real numbers. The limit for \(f(x)\) as \(x\) approaches \(c\) is \(L\) if and only if $$\lim_{x \to c^-} f(x)= L \:\: \text{and}\:\: \lim_{x \to c^+} f(x)= L.$$ The one-sided limit concept extends the continuity definition to closed intervals. Basically, a function is continuous on a closed interval when it is continuous on the interval's interior and exhibits one-sided continuity at the endpoints. This is stated formally in the next definition. Definition 1.4.2 Continuity on a Closed Interval A function \(f\) is continuous on the closed interval \(\mathbf{[a,b]}\) when \(f\) is continuous on the open interval \((a,b)\) and $$\lim_{x \to a^+} f(x)=f(a)$$ and $$\lim_{x \to b^-} f(x)=f(b).$$ The function \(f\) is continuous from the right at \(a\) and continuous from the left at \(b\), as shown in Figure 1.4.7 Similar definitions can be written to cover continuity on intervals with the form \((a,b]\) and \([a,b)\) that are neither open nor closed, or on infinite intervals. For example, \(f(x)=\sqrt{x}\) is continuous on the infinite interval \([0,\infty)\), and the function \(g(x)=\sqrt{2-x}\) is continuous on the infinite interval \((-\infty,2]\).

Example 1.4.4 Continuity on a Closed Interval

\(f\) is continuous on \([-1,1]\). Figure 1.4.8

Describe the continuity for \(f(x)=\sqrt{1-x^2}\). Solution The domain for \(f\) is the closed interval \([-1,1]\). At all points in the open interval \((-1,1)\), the continuity for \(f\) follows from Theorems 1.3.4 and 1.3.5. Moreover, because $$\lim_{x \to -1^+} \sqrt{1-x^2} = 0= f(-1)$$     Continuous from the right and $$\lim_{x \to -1^-} \sqrt{1-x^2} = 0= f(1)$$     Continuous from the left conclude that \(f\) is continuous on the closed interval \([-1,1]\), as shown in Figure 1.4.8.

Example 1.4.5 Charles's Law and Absolute Zero

The hydrogen gas volume depends on its temperature. Figure 1.4.9

On the Kelvin temperature scale absolute zero is the temperature 0 K. The lowest temperature possible. So cold that all motion, even at the atomic level, stops. How did scientists determine that 0 K is the lowest limit for temperature? What is absolute zero on the Celsius scale? Solution The determination for absolute zero stems from the work by the French physicist Jacques Charles (1746–1823). Charles discovered that the volume of gas at a constant pressure increases linearly with the gas's temperature. Table 1.4.1 illustrates this relationship between volume and temperature. To generate the values in the table, one mole of hydrogen gas is held at a constant one atmosphere in pressure. The volume \(V\) is approximated in liters, and the temperature \(T\) is measured in degrees Celsius. Table 1.4.1 Volume vs. Temperature in Celsius The points represented by Table 1.4.1 are shown in Figure 1.4.9 as a graph. By using both the relationship between the variables \(T\) and \(V\) are described by the linear equation $$V=0.08213T+22.4334.$$ Solving for \(T\) produces the equation for the gas's temperature. $$T=\frac{V-22.4334}{0.08213}$$ By reasoning that the gas's volume can approach 0, but can never equal or go below 0, calculating that the least possible temperature is $$\lim_{V \to 0^+} T$$ $$=\lim_{V \to 0^+} \frac{V-22.4334}{0.08213}$$     Using direct substitution. $$=\frac{0-22.4334}{0.08213}$$ \(\approx -273.15. \) Therefore, absolute zero on the Kelvin scale (0K) is approximately -273.15 Celsius°. Table 1.4.2 below shows the temperatures in Example 1.4.5 converted to the Fahrenheit scale. Try repeating the solution shown in Example 1.4.5 using these temperatures and volumes. Use the result to find the value of absolute zero on the Fahrenheit scale. Table 1.4.2 Volume vs. Temperature in Fahrenheit

Continuity Properties

Section 1.3 described limits and their properties. Each property yields a corresponding property pertaining to a function's continuity. For instance, Theorem 1.4.2 follows directly from Theorem 1.3.2.

Theorem 1.4.2 Continuity Properties

If \(b\) is a real number and \(f\) and \(g\) are continuous functions at \(x=c\), then the functions listed below are also continuous at \(c\). 1. Scalar multiple: \(bf\)     2. Sum or difference: \(f \pm g\) 3. Product: \(fg\)     4.Quotient: \(\frac{f}{g},\:\:\:g(c)\ne 0\)

It is important to recognize functions that are continuous at every point in their domains. The list below summarizes the functions covered in this discussion that are continuous at every point in their domains.

1. Polynomial:	\(p(x)=a_{n} x^n+a_{n-1} x^{n-1}+...+a_{1}x+a_{0}\)
2. Relational:	\(r(x)= \frac{p(x)}{q(x)}, \:\:\:\: q(x) \neq 0\)
3. Radical:	\(f(x)=\sqrt[n]{x}\)
4. Trigonometric:	\(\sin x, \cos x, \tan x, \cot x, \sec x, \csc x\)

Applying Theorem 1.4.2 with this list reveals many elementary functions are continuous at every point in their domains.

Example 1.4.6 Applying Continuity Properties

By Theorem 1.4.2, it follows that each function below is continuous at every point in its domain.

$$f(x)=x + \sin x,\:\:\:\:f(x)=3 \tan x,\:\:\:\:f(x)=\frac{x^2+1}{\cos x}$$

Theorem 1.4.3, which follows from Theorem 1.3.5, determines the continuity for composite functions such as

$$f(x)=\sin 3x,\:\:\:\:f(x)=\sqrt{x^2+1},\:\:\:\text{and}\:\:\:f(x)=\tan \frac{1}{x}.$$

Theorem 1.4.3 Continuity for a Composite Function

If \(g\) is continuous at \(c\) and \(f\) is continuous at \(g(c)\), then the composite function given by \((f \circ g)(x)=f(g(x))\) is continuous at \(c\). One consequence is the limit for \(f(g(x))\) as \(x\) approaches \(c\) is $$\lim_{x \to c} f(g(x))=f(g(c))$$

Proof By the definition for continuity, \(\lim_{x \to c}g(x)=g(c)\) and \(\lim_{x \to g(c)}f(x)=f(g(c))\). Apply Theorem 1.3.5 with \(L=g(c)\) to obtain

$$\lim_{x \to c} f(g(x))=f \left ( \lim_{x \to c} g(x) \right)=f(g(c)).$$

Therefore, \((f \circ g)(x)=f(g(x))\) is continuous at \(c\).

Example 1.4.7 Testing for Continuity

Describe the intervals(s) on which each function is continuous.

$$\mathbf{\text{a.}}\:f(x)=\tan x\:\:\:\: \mathbf{\text{b.}} \: g(x)=\left\{\begin{matrix} \sin \frac{1}{x}, & x \neq 0\\ 0,& x=0 \end{matrix}\right. \:\:\:\: \mathbf{\text{c.}}\:h(x)=\left\{\begin{matrix} x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0 \end{matrix}\right.$$

Solution

a.	The tangent \(f(x)= \tan x\) is undefined at $$x=\frac{\pi}{2}+n \pi, \:\:\:\: n\:\text{is an integer}$$ At all other points, \(f\) is continuous. So, \(f(x) = \tan x\) is continuous on the open intervals $$..., \left ( -\frac{3 \pi}{2},-\frac{\pi}{2} \right ),\left ( -\frac{ \pi}{2},-\frac{\pi}{2} \right ),\left ( -\frac{ \pi}{2},-\frac{3\pi}{2} \right ),...$$ As shown in Figure 1.4.10(a)
b.	Because \(y=1/x\) is continuous except at \(x=0\) and the sine function is continuous for all real values for \(x\), if follows from Theorem 1.4.3 that $$y= \sin \frac{1}{x}$$ is continuous at all real values except \(x=0\). At \(x=0\), the limit for \(g(x)\) does not exist (see Section 1.2 Example 1.2.5). Thus, \(g\) is continuous on the intervals \((-\infty,0)\) and \((0,\infty)\), as shown in Figure 1.4.10(b).
c.	This function is similar to the function in part (b) except that the oscillations are damped by the factor \(x\). Using the Squeeze Theorem obtains $$-|x| \leq x \sin \frac{1}{x} \leq |x|,\:\:x \neq 0$$ and conclude that $$\lim_{x \to 0} h(x)=0.$$ Therefore \(h\) is continuous on the entire real number line, as shown in Figure 1.4.10(c)

The Intermediate Value Theorem

\(f\) is continuous on \([a,b]\). [There exist three \(c\)'s such that \(f(c) =k\).] Figure 1.4.11 \(f\) is not continuous on \([a,b]\). [There are no \(c\)'s such that \(f(c) =k\).] Figure 1.4.12

Theorem 1.4.4 The Intermediate Value Theorem If \(f\) is continuous on the closed interval \([a,b],\:f(a) \neq f(b)\), and \(k\) is any number between \(f(x)\) and \(f(b)\), then there is at least one number \(c\) in \([a,b]\) such that \(f(c) =k\). The Intermediate Value Theorem states that at least one number \(c\) exists, but it does not provide a method for finding \(c\). Such theorems are called existence theorems. The proof for this theorem is based on a property for real numbers called completeness. The Intermediate Value Theorem states that for a continuous function \(f\), if \(x\) takes on all values between \(a\) and \(b\), then \(f(x)\) must take on all values between \(f(a)\) and \(f(b)\). For an example, consider a person’s height. A girl is 5 feet tall on her thirteenth birthday and 5 feet 7 inches tall on her fourteenth birthday. Then, for any height \(h\) between 5 feet and 5 feet 7 inches, there must have been a time \(t\) when her height was exactly \(h\). This seems reasonable because human growth is continuous and a person’s height does not abruptly change from one value to another. The Intermediate Value Theorem guarantees that at least one number c exists in the closed interval \([a,b]\). There may be more than one number \(c\) such that \(f(c)=k\) as shown in Figure 1.4.11. A function that is not continuous does not necessarily exhibit the intermediate value property. For example, the graph for the function shown in Figure 1.4.12 jumps over the horizontal line \(y=k\) and for this function there is no value for \(c\) in \([a,b]\) such that \(f(c)=k\). The Intermediate Value Theorem often can be used to locate a function's zeros as long as it is continuous on a closed interval. Specifically, if \(f\) is continuous on \([a,b]\) and \(f(a)\) and \(f(b)\) differ in sign, then the Intermediate Value Theorem guarantees the existence of at least one zero of f in the closed interval \([a,b]\).

Example 1.4.8 Applying the Intermediate Value Theorem

Th Figure 1.4.13

Use the Intermediate Value Theorem to show that the polynomial function \(f(x)=x^3+2x-1\) has a zero in the interval \([0,1]\). Solution Note that \(f\) is continuous on the closed interval \([0,1]\). Because \(f({\color{Red} x} )={\color{Red} 0}^3+2({\color{Red} 0})-1=-1\:\:\text{ and }\:\: f({\color{Red} 1})={\color{Red} 1}^3+2({\color{Red} 1})-1=2\) it follows that \(f(0)<0\) and \(f(1)>0\). Therefore the Intermediate Value Theorem can be applied to conclude that there must be some \(c\) in \([0,1]\) such that \(f(c)=0\:\:\:\: {\color{Red} f}\) has a zero in the closed interval [0,1]. as shown in Figure 1.4.13. The bisection method for approximating the real zeros for a continuous function is similar to the method used in Example 1.4.8. If zero is known in the closed interval \([a,b]\), then the zero must lie in the interval \([a,(a+b)/2]\) or \([(a+b)/2,b]\). From the sign for \(f(a+b)/2)\), the interval containing the zero can be determined. Repeatedly bisect the interval until the zero is revealed.

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Parent Article: Calculus I 01 Limits and Their Properties