Combinations of Continuous Functions

Section 3.1 Problems

3.2 Continuity

3.2.2 Combinations of Continuous Functions

EXAMPLE 5 At which point is the function

f(x)= ¹ (x−⁴)² discontinuous? Can the discontinuity be removed?

Solution

The graph of f(x)is shown in Figure 3.14. The function f(x)cannot be defined for x =^{4, since} f(x)is of the form¹₀ whenx =4. The function is defined for all other values ofx. Therefore, we look atx =4. We find that

lim

x→4

1

(x−4)² = ∞ (limit does not exist)

Because∞is not a real number, we cannot assign a value to f(⁴)^{such that} f(x) would be continuous atx =4. We therefore conclude that f(x)is discontinuous at x =4 and the discontinuity cannot be removed.

4 5 6 7 8 x

3

0 1 2

30 20 10 40 y

y

Figure 3.14 The function f(x)=_(x−¹₄₎2is discontinuous atx=4.

3. Since f andgare continuous atx =c, it follows that lim

x→c

f(x)= f(c) ^{and lim}

x→c

g(x)=g(c) ^(3.6) Therefore, combining (3.5) and (3.6), we obtain

lim

x→c[f +g](x)= ^lim

x→c

f(x)+^lim

x→c

g(x)= f(c)+g(c) which is equal to[f +g](c)and hence condition 3 holds.

Since we showed that all three conditions hold, it follows that f+gis continuous atx=c. The other statements are shown in a similar way, using the limit laws.

We say that a function f is continuous on an intervalI if f is continuous for all x ∈I. Note that ifI is a closed interval, then continuity at the left (and, respectively, right) endpoint of the interval means continuous from the right (and, respectively, left). Many of the elementary functions are indeed continuous wherever they are defined. For polynomials and rational functions, this statement follows immediately from the fact that certain combinations of continuous functions are continuous. We give a list of the most important cases:

The following functions are continuous wherever they are defined:

1. polynomial functions 2. rational functions 3. power functions 4. trigonometric functions

5. exponential functions of the forma^x,a>^{0 and}a=¹ 6. logarithmic functions of the form log_ax,a>^{0 and}a=¹

The phrase “wherever they are defined” is crucial. It helps us to identify points where a function might be discontinuous. For instance, the power function 1/x² is defined only forx=0, and the logarithmic function log_axis defined only forx >^0.

We will illustrate the six cases cited in the preceding box in the next example, paying particular attention to the phrase “wherever they are defined.”

EXAMPLE 6 For which values ofx ∈^Rare the following functions continuous?

(a) f(x)=²x³−³x+^{1 (b)} f(x)= x²+x+¹

x−^{2 (c)} f(x)=x^1/4 (d) f(x)=3 sinx (e) f(x)=tanx (f) f(x)=3^x (g) 2 ln(x+¹)

Solution

(a) f(x)is a polynomial and is defined for allx ∈ R; it is therefore continuous for allx∈R.

(b) f(x)is a rational function defined for allx =2; it is therefore continuous for allx =^2.

(c) f(x) = x^1/4 = √⁴

x is a power function defined forx ≥ 0; it is therefore continuous forx ≥^0.

(d) f(x)is a trigonometric function. Because sinxis defined for allx∈^{R, 3 sin}x is continuous for allx∈^R.

(e) f(x)is a trigonometric function. The tangent function is defined for allx =

π

2 +kπ^{, where}kis an integer; it is therefore continuous for allx = ^π₂ +kπ^{, where}k is an integer.

(f) f(x)is an exponential function. f(x) = ^3x is defined for allx ∈ ^{Rand is} therefore continuous for allx ∈^R.

(g) f(x)is a logarithmic function.f(x)=^{2 ln}(x+¹)is defined as long asx+¹>⁰ orx >−1; it is therefore continuous for allx>−^1.

The following result is useful in determining whether a composition of functions is continuous:

Theorem Ifg(x)is continuous atx=cwithg(c)=Land f(x)is continuous atx = L, then(f ◦g)(x)is continuous atx =c. In particular,

lim

x→c(f ◦g)(x)= ^lim

x→c

f[g(x)] = f[^lim

x→c

g(x)] = f[g(c)] = f(L)

To explain this theorem, recall what it means to compute(f ◦g)(c)= f[g(c)]. When we compute f[g(c)], we take the valuec, compute g(c), and then take the resultg(c)and plug it into the function f to obtain f[g(c)]. If, at each step, the functions are continuous, the resulting function will be continuous.

EXAMPLE 7 Determine where the following functions are continuous:

(a) h(x)=e^−x2 (b) h(x)=^sinπ

x ^(c) h(x)= ¹ 1+²x^1/3

Solution

(a) Set g(x) = −x² and f(x) = e^x. Thenh(x) = (f ◦g)(x)^{. Since} g(x)^{is a} polynomial, it is continuous for allx ∈ R, and the range of g(x)^is(−∞,⁰]^. f(x) is continuous for all values in the range ofg(x). [In fact, f(x)is continuous for all x ∈R.] It therefore follows thath(x)is continuous for allx∈^R.

(b) Setg(x)= ^π_x ^and f(x)= ^sinx.g(x)is continuous for allx = 0. The range ofg(x)is the set of all real numbers, excluding 0. f(x)is continuous for allx in the range ofg(x). Hence, h(x)is continuous for allx = 0. Recall that we showed in Example 8 of Section 3.1 that

lim

x→0sinπ x does not exist. That is,h(x)is discontinuous atx =^0.

(c) Setg(x)=x^1/3and f(x)= _1+2x^{1 . Then}h(x)=(f ◦g)(x)^.g(x)is continuous for allx ∈^{R, since}g(x)= x^1/3 = √³

x and 3 is an odd integer. The range ofg(x)^is (−∞,∞)^. f(x)is continuous for all realxdifferent from−¹/^{2. Since}g(−¹₈)= −¹₂^, h(x)is continuous for all realxdifferent from−¹/8. Another way to see that we need to exclude−¹₈ from the domain ofh(x)is by looking directly at the denominator of h(x). We have 1+²x^1/3=^{0 when}x= −¹₈^.

When we compute lim_x→c f(x)and we know that f(x)is continuous atx =c, it follows that lim_x→c f(x)= f(c). The next three examples illustrate this property.

EXAMPLE 8 Find

lim

x→3sin

πx²−¹ 4

Solution

The function f(x)=^sin π^x2₄⁻¹

is continuous atx =^{3. Hence,}

lim

x→3sin

πx²−¹ 4

=^sin

π⁹−¹ 4

=^sin(²π)=⁰

EXAMPLE 9 Find

lim

x→1

2x³−1

Solution

The function f(x)=

2x³−1 is continuous atx =^{1. Thus,} lim

x→1

2x³−¹=

(²)(¹)³−¹= 1=¹

EXAMPLE 10 Find

lim

x→0e^x−1

Solution

The function f(x)=e^x−1is continuous atx =0. Therefore, lim

x→0e^x−1=e⁰⁻¹=e⁻¹

We conclude this section by calculating the limit of the expression in Example 9 of Section 3.1.

EXAMPLE 11 Find

lim

x→0

x²+¹⁶−⁴ x²

Solution

We cannot apply Rule 4 of Section 3.1, since f(x)=(

x²+¹⁶−⁴)/x²is not defined forx =0. (If we plug in 0, we get the expression 0/0.) We use a trick that will allow us to find the limit: We rationalize the numerator. Forx =0, we find that

x²+16−4

x² = (

x²+16−4) x²

(

x²+16+4) (

x²+¹⁶+⁴)

= x²+¹⁶−¹⁶ x²(

x²+¹⁶+⁴) = x² x²(

x²+¹⁶+⁴)

= ¹ x²+¹⁶+⁴

Note that we are allowed to divide byx²in the last step, since we are assuming thatx =0. We can now apply Rule 4 to 1/(

x²+¹⁶+⁴). When we do, we obtain lim

x→0

x²+¹⁶−⁴ x² = ^lim

x→0

1

x²+16+4 = ¹

8 =⁰.¹²⁵

as we saw in Example 9 of Section 3.1. In Chapter 5, we will learn another method for finding the limit of expressions of the form 0/^0.