4.8 Continuity - Avidemia

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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 $x = - 3, - 1, 3, 5$ .

Figure 1: The function

f

is not continuous at

x = - 3, - 1, 3

and

5

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

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

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

Figure 2

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 $f (a) .$

A function that is discontinuous at every point of its domain

Consider the Dirichlet function defined as $D (x) = {\begin{cases} 1 & if x is rational \\ 0 & if x is irrational \end{cases}$ 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 $ϵ = 1 / 2$ is given, we should be able to find a $δ > 0$ such that $0 < | x - a | < δ ⟹ | D (x) - L | < \frac{1}{2}$ Because each deleted neighborhood $0 < | x - a | < δ$ 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

Power functions. $y = x^{n}$ where $n$ is a positive integer is continuous everywhere.
Polynomials are continuous everywhere, because for any $a$ , $lim_{x \to a} P (x) = P (a)$
Rational functions. 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.
Root functions (1). $y = \sqrt[n]{x}$ where $n$ is a positive odd integer is continuous everywhere.
Root functions (2). $y = \sqrt[n]{x}$ where $n$ is a positive even integer is continuous on its domain $[0, \infty)$ .
The sine and cosine functions are continuous on $R = (- \infty, \infty)$ .
Figure 8: $y = \sin x$ and $y = \cos x$ are continuous everywhere.
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 $x = \frac{π}{2} + k π$ for all integers $k$ . Thus $y = \tan x$ is continuous on ${x | x \neq \frac{π}{2} + k π, k \in Z}$

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) $y = csc x$

Figure 10
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 $R = (- \infty, \infty)$ .
Exponential functions. $y = e^{x}$ is continuous on its domain $R = (- \infty, \infty)$ .
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:

The greatest integer function (also known as the floor function) $f (x) = ⌊ x ⌋$ (or $y = [[x]]$ ) which is discontinuous at every integer (although it is defined everywhere).

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

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

Figure 11

Example 2

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

Solution

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

Example 3

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

Solution

$f$ is a rational function, so it is discontinuous at the points where the denominator $x^{2} - 3 x + 2$ is zero: $x^{2} - 3 x + 2 = 0 \Rightarrow x = \frac{3 \pm \sqrt{3^{2} - 4 (2)}}{2 (1)}$ [Recall that if $a x^{2} + b x + c = 0$ , then $x = (- b \pm \sqrt{b^{2} - 4 a c}) / (2 a)$ ] $\Rightarrow x = 2 or x = 1$ Therefore, $f$ is discontinuous at $x = 1$ and $x = 2$ . However, $f$ has a removable discontinuity at $x = 2$ , because $\frac{x^{2} - 4}{x^{2} - 3 x + 2} = \frac{(x - 2) (x + 2)}{(x - 1) (x - 2)} = \frac{x + 2}{x - 1} (if x \neq 2) .$ [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 $g (x) = {\begin{cases} \frac{x^{2} - 4}{x^{2} - 3 x + 2} & if x \neq 2 \\ 4 & if x = 2 \end{cases}$ 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} - 3 x + 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 $k f (x)$ 4. Product $f (x) g (x)$ 5. Quotient $\frac{f (x)}{g (x)} provided g (a) \neq 0$ 6. Power ${(f (x))}^{n} n is a positive integer$

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 $ϵ > 0$ be given. We need to find a number $δ > 0$ such that for all $x$ $0 < | x - a | < δ \Rightarrow | f (g (x)) - f (b) | < ϵ .$ Because $f$ is continuous at $b$ , we have $lim_{y \to b} f (y) = f (b)$ and therefore, there exists a number $δ_{1} > 0$ such that for all $y$ $| y - b | < δ_{1} \Rightarrow | f (y) - f (b) | < ϵ$ [Because $f$ is continuous, we did not exclude $y = b$ and wrote $| y - b | < δ_{1}$ instead of $0 < | y - b | < δ_{1}$ ].
Because $lim_{x \to a} g (x) = b$ , there exists a $δ > 0$ such that for all $x$ $0 < | x - a | < δ \Rightarrow | g (x) - b | < δ_{1}$ If we let $y = g (x)$ , we can combine these two statements: For all $x$ if $0 < | x - a | < δ$ then $| g (x) - b | < δ_{1}$ which implies that $| f (g (x)) - f (b) | < ϵ$ . 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 $lim_{x \to a} f (g (x)) = f (lim_{x \to a} g (x))$ 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) $lim_{x \to - 3} \sin (\frac{x + 3}{x^{2} + 3 x})$
(b) $lim_{x \to 1} \arctan (x + \cos (\frac{π}{2} x))$
(c) $lim_{x \to 0} \ln (1 + e^{\tan x})$

Solution

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

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

(c) Because $y = \ln x$ is a continuous function on its domain $(0, \infty)$ , we have $\begin{aligned} lim_{x \to 0} \ln (1 + e^{\tan x}) & = \ln (lim_{x \to 0} (1 + e^{\tan x})) \\ = \ln (lim_{x \to 0} 1 + lim_{x \to 0} e^{\tan x}) \end{aligned}$ Because $y = e^{x}$ is continuous everywhere, $\begin{aligned} \ln (lim_{x \to 0} 1 + lim_{x \to 0} e^{\tan x}) & = \ln (lim_{x \to 0} 1 + e^{lim_{x \to 0} \tan x}) \\ = \ln (1 + e^{0}) \\ = \ln (1 + 1) = \ln 2. \end{aligned}$ That is, $lim_{x \to 0} \ln (1 + e^{\tan x}) = \ln 2.$

Example 5

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

Solution

Because $f (x) = \sin x$ is continuous everywhere, $lim_{x \to \infty} \sin \frac{1}{x} = \sin (lim_{x \to \infty} \frac{1}{x}) .$ [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 $\sin (lim_{x \to \infty} \frac{1}{x}) = \sin (0) = 0.$ 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) $lim_{x \to 0^{+}} \arctan (\frac{1}{x})$
(b) $lim_{x \to 0^{-}} \arctan (\frac{1}{x})$
(c) $lim_{x \to \infty} \arctan (\frac{1}{x})$
(d) $lim_{x \to - \infty} \arctan (\frac{1}{x})$

Solution

(a) Because $y = \arctan x$ is a continuous function, we can find the limit of its input first. That is, $\begin{aligned} lim_{x \to 0^{+}} \arctan (\frac{1}{x}) & = \arctan (lim_{x \to 0^{+}} \frac{1}{x}) \\ = \arctan (+ \infty) \\ = \frac{π}{2} \end{aligned}$ [To be accurate, we cannot write $\arctan (+ \infty)$ because $+ \infty$ is not a number] (b) Similar to part (a) $\begin{aligned} lim_{x \to 0^{-}} \arctan (\frac{1}{x}) & = \arctan (lim_{x \to 0^{-}} \frac{1}{x}) \\ = \arctan (- \infty) \\ = - \frac{π}{2} . \end{aligned}$

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

(d) $\begin{aligned} lim_{x \to - \infty} \arctan (\frac{1}{x}) & = \arctan (lim_{x \to - \infty} \frac{1}{x}) \\ = \arctan (0) \\ = 0. \end{aligned}$ 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 Theorem 2 $\begin{aligned} lim_{x \to a} f (g (x)) & = f (lim_{x \to a} g (x)) \\ = f (g (a)) . \end{aligned}$

Example 7

Where are the following functions continuous?
(a)

h (x) = \cos (x^{3} + 1)

(b)

F (x) = | \frac{x \cos x}{x^{2} + 1} |

(c)

G (x) = \sqrt{x^{2} + 2 x - 2}

Solution

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

(b) Notice that $F$ is the composition of the following functions $F = f \circ g$ : $f (x) = | x | and g (x) = \frac{x \cos x}{x^{2} + 1}$ Obviously $f$ is continuous on $R$ . 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 $R$ and $w (x) \neq 0$ , by the product and quotient rules (Theorem 1), $g$ is continuous on $R$ too. Because $f$ and $g$ are both continuous on $R$ and $F = f \circ g$ , it follows from Theorem 3 that $F$ is also continuous on $R$ .

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

$\large x$		$- 1 - \sqrt{3}$		$- 1 + \sqrt{3}$
sign of $(x + 1 - \sqrt{3})$	– – – –	–	– – – –	$0$	+ + + +
sign of $(x + 1 + \sqrt{3})$	– – – –	$0$	+ + + +	+	+ + + +
$∴$ sign of $x^{2} + 2 x - 2$	+ + + +	$0$	– – – –	$0$	+ + + +

Therefore, $G$ is continuous on the following set: ${x \in R | x^{2} + 2 x - 2 \geq 0} = (- \infty, - 1 - \sqrt{3}] \cup [- 1 + \sqrt{3}, \infty)$

Definitions of Continuity and Discontinuity

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Left and Right Continuities

Continuity on an Interval

Types of Discontinuity

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Read More on Jump Discontinuity

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Read More on Infinite Discontinuity

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Read More on Oscillating Discontinuity

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A function that is discontinuous at every point of its domain

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Elementary Continuous Functions

Algebraic Operations on Continuous Functions

Continuity of Composite Functions

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Proof of the Root Rule from Sec. 4.4

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