[Calculus] II. Continuity

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Continuity

Definition 2.1.1.  A function \( f \) is continuous at a number \( a \) if

\( \displaystyle \lim_{x \to a} f(x) = f(a) \)

Remark 2.1.2.  Notice that Definition 2.1.1 implicitly requires three things if \( f \) is continuous at \( a \):

\( \displaystyle (1) \hspace{2.5em} f(a) \) is defined.

\( \displaystyle (2) \hspace{2.5em} \lim_{x \to a}f(x) \) exists.

\( \displaystyle (3) \hspace{2.5em} \lim_{x \to a}f(x) = f(a) \)

The definition says that \( f \) is continuous at \( a \) if \( f(x) \) approaches \( f(a) \) as \( x \) approaches \( a \). Thus a continuous function \( f \) has the property that a small change in \( x \) produces only a small change in \( f(x) \). In fact, the change in \( f(x) \) can be kept as small as we please by keeping the change in \( x \) sufficiently small.

If \( f \) is defined near \( a \) (in other words, \( f \) is defined on an open interval containing \( a \), except perhaps at \( a \)), we say that \( f \) is discontinuous at \( a \) (or \( f \) has a discontinuity at \( a \)) if \( f \) is not continuous at \( a \).

Definition 2.1.3.  A function \( f \) is continuous from the right at a number \( a \) if

\( \displaystyle \lim_{x \to a^{+}}\!f(x) = f(a) \)

and \( f \) is continuous from the left at \( a \) if

\( \displaystyle \lim_{x \to a^{-}}\!f(x) = f(a) \)

Definition 2.1.4.  A function \( f \) is continuous on an interval if it is continuous at every number in the interval. (If \( f \) is defined only on one side of an endpoint of the interval, we understand continuous at the endpoint to mean continuous from the right or continuous from the left.)

Example 2.1.5.  Show that the function \( f(x) = 1 - \sqrt{1 - x^{2}} \) is continuous on the interval \( [-1,\ 1] \).

Solution.  If \( -1 < a < 1 \), then we have

\( \displaystyle \begin{align} \lim_{x \to a}f(x) &= \lim_{x \to a}(1 - \sqrt{1 - x^{2}}) \\[7pt] &= 1 - \lim_{x \to a}\sqrt{1 - x^{2}} \\[7pt] &= 1 - \sqrt{\lim_{x \to a}(1 - x^{2})} \\[7pt] &= 1 - \sqrt{1 - a^{2}} \\[7pt] &= f(a) \end{align} \)

Thus, by Definition 2.1.1, \( f \) is continuous at \( a \) if \( -1 < a < 1 \). Similar calculations show that

\( \displaystyle \lim_{x \to {-1}^{+}}\!\!f(x) = 1 = f(-1) \quad \text{and} \quad \lim_{x \to {1}^{-}}\!f(x) = 1 = f(1) \)

so \( f \) is continuous from the right at \( -1 \) and continuous from the left at \( 1 \). Therefore, according to Definition 2.1.4, \( f \) is continuous on \( [-1,\ 1] \).

\( \Box \)

Theorem 2.1.6. (Limit Laws)  If \( f \) and \( g \) are continuous at \( a \) and if \( c \) is a constant, then the following functions are also continuous at \( a \):

\( \displaystyle (1) \hspace{2.5em} f + g \)

\( \displaystyle (2) \hspace{2.5em} f - g \)

\( \displaystyle (3) \hspace{2.5em} cf \)

\( \displaystyle (4) \hspace{2.5em} fg \)

\( \displaystyle (5) \hspace{2.5em} \frac{f}{g} \quad \text{if} \ \ g(a) \neq 0 \)

Proof.  Each of the five parts of this theorem follows from the corresponding Theorem 1.2.1. For instance, we give the proof of \( \displaystyle (1) \). Since \( f \) and \( g \) are continuous at \( a \), we have

\( \displaystyle \lim_{x \to a}f(x) = f(a) \quad \text{and} \quad \lim_{x \to a}g(x) = g(a) \)

Therefore

\( \displaystyle \begin{align} \lim_{x \to a}(f + g)(x) &= \lim_{x \to a}[f(x) + g(x)] \\[7pt] &= \lim_{x \to a}f(x) + \lim_{x \to a}g(x) \\[7pt] &= f(a) + g(a) \\[7pt] &= (f + g)(a) \end{align} \)

This shows that \( f + g \) is continuous at \( a \).

\( \Box \)

Theorem 2.1.7.  Any polynomial is continuous everywhere; that is, it is continuous on \( \mathbb{R} = (-\infty,\ \infty) \). Also, any rational function is continuous wherever it is defined; that is, it is continuous on its domain.

Proof.  A polynomial is a function of the form

\( \displaystyle P(x) = c_{n}x^{n} + c_{n - 1}x^{n - 1} + \cdots + c_{1}x + c_{0} \)

where \( c_{0}, c_{1}, \cdots , c_{n} \) are constants. We know that

\( \displaystyle \lim_{x \to a}c_{0} = c_{0} \)

and

\( \displaystyle \lim_{x \to a}x^{m} = a^{m} \qquad m = 1, 2, \cdots , n \)

This equation is precisely the statement that the function \( f(x) = x^{m} \) is a continuous function. Thus, by Theorem 2.1.6 - \( (3) \), the function \( g(x) = cx^{m} \) is continuous. Since \( P \) is a sum of functions of this form and a constant function, it follows from Theorem 2.1.6 - \( (1) \) that \( P \) is continuous.

A rational function is a function of the form

\( \displaystyle f(x) = \frac{P(x)}{Q(x)} \)

where \( P \) and \( Q \) are polynomials. The domain of \( f \) is \( D = \{ x \in \mathbb{R}\ |\ Q(x) \neq 0 \} \). We know that \( P \) and \( Q \) are continuous everywhere. Thus, by Theorem 2.1.6 - \( (5) \), \( f \) is continuous at every number in \( D \).

\( \Box \)

Theorem 2.1.8.  Polynomials, root functions, rational functions, and trigonometric functions are continuous at every number in their domains.

Example 2.1.9.  Evaluate \( \displaystyle \lim_{x \to \pi} \frac{\sin x}{2 + \cos x} \).

Solution.  Theorem 2.1.8 tells us that \( y = \sin x \) is continuous. The function in the denominator, \( y = 2 + \cos x \), is the sum of two continuous functions and is therefore continuous. Notice that this function is never \( 0 \) because \( \cos x \ge -1 \) for all \( x \) and so \( 2 + \cos x > 0 \) everywhere. Thus the ratio

\( \displaystyle f(x) = \frac{\sin x}{2 + \cos x} \)

is continuous everywhere. Hence, by the definition of a continuous function,

\( \displaystyle \lim_{x \to \pi}\frac{\sin x}{2 + \cos x} = \lim_{x \to \pi}f(x) = f(\pi) = \frac{\sin \pi}{2 + \cos \pi} = \frac{0}{2 - 1} = 0 \)

\( \Box \)

Theorem 2.1.10.  If \( f \) is continuous at \( b \) and \( \displaystyle \lim_{x \to a}g(x) = b \), then \( \displaystyle \lim_{x \to a}f(g(x)) = f(g(a)) = f(b) \). In other words,

\( \displaystyle \lim_{x \to a}f(g(x)) = f \! \left(\lim_{x \to a}g(x) \right) \)

Proof.  Let \( \varepsilon > 0 \) be given. We want to find a number \( \delta > 0 \) such that

\( \displaystyle \text{if} \quad 0 < |\ x - a\ | < \delta \quad \text{then} \quad \left| \ f(g(x)) - f(b)\ \right| < \varepsilon \)

Since \( f \) is continuous at \( b \), we have

\( \displaystyle \lim_{y \to b}f(y) =f(b) \)

and so there exists \( \delta_{1} > 0 \) such that

\( \displaystyle \text{if} \quad 0 < |\ y - b\ | < \delta_{1} \quad \text{then} \quad \left| \ f(y) - f(b)\ \right| < \varepsilon \)

Since \( \displaystyle \lim_{x \to a}g(x) = b \), there exists \( \delta > 0 \) such that

\( \displaystyle \text{if} \quad 0 < |\ x - a\ | < \delta \quad \text{then} \quad \left| \ g(x) - b\ \right| < \delta_{1} \)

Combining these two statements, we see that whenever \( 0 < |\ x - a\ | < \delta \) we have \( |\ g(x) - b\ | < \delta_{1} \), which implies that \( \left|\ f(g(x)) - f(b)\ \right| < \varepsilon \). Therefore we have proved that \( \displaystyle \lim_{x \to a}f(g(x)) = f(b) \).

\( \Box \)

Theorem 2.1.11.  If \( g \) is continuous at \( a \) and \( f \) is continuous at \( g(a) \), then the composite function \( f \circ g \) given by \( (f \circ g)(x) = f(g(x)) \) is continuous at \( a \).

Proof.  Since \( g \) is continuous at \( a \), we have

\( \displaystyle \lim_{x \to a}g(x) = g(a) \)

Since \( f \) is continuous at \( b = g(a) \), we can apply Theorem 2.1.10 to obtain

\( \displaystyle \lim_{x \to a}f(g(x)) = f(g(a)) \)

which is precisely the statement that the function \( h(x) = f(g(x)) \) is continuous at \( a \); that is, \( f \circ g \) is continuous at \( a \).

\( \Box \)

Theorem 2.1.12. (The Intermediate Value Theorem)  Suppose that \( f \) is continuous on the closed interval \( [a,\ b] \) and let \( N \) be any number between \( f(a) \) and \( f(b) \), where \( f(a) \neq f(b) \). Then there exists a number \( c \) in \( (a,\ b) \) such that \( f(c) = N \).

Remark 2.1.13.  Theorem 2.1.12 states that a continuous function takes on every intermediate value between the function values \( f(a) \) and \( f(b) \). It is illustrated by Figure 1. Note that the value \( N \) can be taken on once or more than once.

Figure 1

Example 2.1.14.  Show that there is a root of the equation

\( \displaystyle 4x^{3} - 6x^{2} + 3x - 2 = 0 \)

between \( 1 \) and \( 2 \).

Solution.  Let \( f(x) = 4x^{3} - 6x^{2} + 3x - 2 \). We are looking for a solution of the given equation, that is, a number \( c \) between \( 1 \) and \( 2 \) such that \( f(c) = 0 \). Therefore we take \( a = 1 \), \( b = 2 \), and \( N = 0 \) in Theorem 2.1.12. We have

\( \displaystyle \begin{align} &f(1) = 4 - 6 + 3 - 2 = -1 < 0 \\[7pt] &f(2) = 32 - 24 + 6 - 2 = 12 > 0 \end{align} \)

Thus \( f(1) < 0 < f(2) \); that is, \( N = 0 \) is a number between \( f(1) \) and \( f(2) \). Now \( f \) is continuous since it is a polynomial, so the Theorem 2.1.12 says there is a number \( c \) between \( 1 \) and \( 2 \) such that \( f(c) = 0 \). In other words, the equation \( 4x^{3} - 6x^{2} + 3x -2 = 0 \) has at least one root \( c \) in the interval \( \left(1,\ 2\right) \).

\( \Box \)

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The body text and the figures were excerpted from James Stewart, Calculus 8E.