1 + ex 3

On multiplying through, we get

1

3ex = 1 + ex or 2ex = 1, ex =

2

x = ln(1/2) = ’ ln(2).

Example 1.4.4 Prove that for all real x, cosh2 x ’ sinh2 x = 1.

2 2

1 1

cosh2 x ’ sinh2 x = (ex + e’x ) ’ (ex ’ e’x )

2 2

1

= [e2x + 2 + e’2x ) ’ (e2x ’ 2 + e’2x )]

4

1

= [4]

4

=1

30 CHAPTER 1. FUNCTIONS

Example 1.4.5 Prove that

(a) sinh(x + y) = sinh x cosh y + cosh x sinh y.

(b) sinh 2x = 2 sinh x cosh y.

Equation (b) follows from equation (a) by letting x = y. So, we work

with equation (a).

1

1

(a) sinh x cosh y + cosh x sinh y = (ex ’ e’x ) · (ey + e’y )

2 2

1 1

+ (ex + e’x ) · (ey ’ e’y )

2 2

1

= [(ex+y + ex’y ’ e’x+y ’ e’x’y )

4

+ (ex+y ’ ex’y + e’x+y ’ e’x’y )]

1

= [2(ex+y ’ e’(x+y) ]

4

1

= (e(x+y) ’ e’(x+y) )

2

= sinh(x + y).

Example 1.4.6 Find the inverses of the following functions:

(a) sinh x (b) cosh x (c) tanh x

1

(a) Let y = sinh x = (ex ’ e’x ). Then

2

1x

(e ’ e’x )

2ex y = 2ex = e2x ’ 1

2

e2x ’ 2yex ’ 1 = 0

(ex )2 ’ (2y)ex ’ 1 = 0

4y 2 + 4

2y ±

x

y2 + 1

=y±

e=

2

Since ex > 0 for all x, ex = y + 1 + y2.

1.4. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS31

On taking natural logarithms of both sides, we get

1 + y 2 ).

x = ln(y +

The inverse function of sinh x, denoted arcsinh x, is de¬ned by

√

arcsinh x = ln(x + 1 + x2 )

(b) As in part (a), we let y = cosh x and

1

2ex y = 2ex · (ex + e’x ) = e2x + 1

2

2x x

e ’ (2y)e + 1 = 0

4y 2 ’ 4

2y ±

x

e=

2

ex = y ± y 2 ’ 1.

We observe that cosh x is an even function and hence it is not one-to-

one. Since cosh(’x) = cosh(x), we will solve for the larger x. On taking

natural logarithms of both sides, we get

y 2 ’ 1) or x2 = ln(y ’ y 2 ’ 1).

x1 = ln(y +

We observe that

y 2 ’ 1)(y + y 2 ’ 1)

(y ’

y2

x2 = ln(y ’ ’ 1) = ln

y2 ’ 1

y+

1

= ln

y2 ’ 1

y+

y 2 ’ 1) = ’x1 .

= ’ ln(y +

Thus, we can de¬ne, as the principal branch,

√

x2 ’ 1), x ≥ 1

arccosh x = ln(x +

32 CHAPTER 1. FUNCTIONS

(c) We begin with y = tanh x and clear denominators to get

ex ’ e’x

|y| < 1

y= x ,

e + e’x

ex [(ex + e’x )y] = ex [(ex ’ e’x )] |y| < 1

,

(e2x + 1)y = e2x ’ 1 |y| < 1

,

e2x (y ’ 1) = ’(1 + y) |y| < 1

,

(1 + y)

e2x = ’ |y| < 1

,

y’1

1+y

e2x = |y| < 1

,

1’y

1+y

|y| < 1

2x = ln ,

1’y

1 1+y

, |y| < 1.

x = ln

1’y

2

Therefore, the inverse of the function tanh x, denoted arctanhx, is de¬ned

by

1 1+x

, |x| < 1.

arctanh , x = ln

1’x

2

Exercises 1.4

1. Evaluate each of the following

(c) ln(e0.001 )

(a) log10 (0.001) (b) log2 (1/64)

0.1

(100)1/3 (0.01)2 ’2 )

(e) eln(e

(d) log10

(.0001)2/3

2. Prove each of the following identities

(a) sinh(x ’ y) = sinh x cosh y ’ cosh x sinh y

(b) cosh(x + y) = cosh x cosh y + sinh x sinh y

1.4. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS33

(c) cosh(x ’ y) = cosh x cosh y ’ sinh x sinh y

(d) cosh 2x = cosh2 x + sinh2 x = 2 cosh2 x ’ 1 = 1 + 2 sinh2 x

3. Simplify the radical expression by using the given substitution.

√ √

2 + x2 , x = a sinh t (b) x2 ’ a2 , x = a cosh t

(a) a

√

(c) a2 ’ x2 , x = a tanh t

4. Find the inverses of the following functions:

(a) coth x (b) sech x (c) csch x

3

5. If cosh x = , ¬nd sinh x and tanh x.

2

6. Prove that sinh(3t) = 3 sinh t + 4 sinh3 t (Hint: Expand sinh(2t + t).)

7. Sketch the graph of each of the following functions.

c) y = 10’x d) y = 2’x

a) y = 10x b) y = 2x

2 2

f) y = e’x g) y = xe’x i) y = e’x

e) y = ex

j) y = sinh x k) y = cosh x l) y = tanh x m) y = coth x

n) y = sech x o) y = csch x

8. Sketch the graph of each of the following functions.

a) y = log10 x b) y = log2 x c) y = ln x d) y = log3 x

e) y = arcsinh x f) y = arccosh x g) y = arctanh x

9. Compute the given logarithms in terms log10 2 and log10 3.

34 CHAPTER 1. FUNCTIONS

27 20

a) log10 36 b) log10 c) log10

16 9

610

30

d) log10 (600) e) log10 f) log10

(20)5

16

10. Solve each of the following equations for the independent variable.

a) ln x ’ ln(x + 1) = ln(4) b) 2 log10 (x ’ 3) = log10 (x + 5) + log10 4

c) log10 t2 = (log10 t)2 d) e2x ’ 4ex + 3 = 0

e) ex + 6e’x = 5 f) 2 sinh x + cosh x = 4

Chapter 2

Limits and Continuity

2.1 Intuitive treatment and de¬nitions

2.1.1 Introductory Examples

The concepts of limit and continuity are very closely related. An intuitive

understanding of these concepts can be obtained through the following ex-

amples.

Example 2.1.1 Consider the function f (x) = x2 as x tends to 2.

As x tends to 2 from the right or from the left, f (x) tends to 4. The

value of f at 2 is 4. The graph of f is in one piece and there are no holes or

jumps in the graph. We say that f is continuous at 2 because f (x) tends to

f (2) as x tends to 2.

graph

The statement that f (x) tends to 4 as x tends to 2 from the right is

expressed in symbols as

lim f (x) = 4

+

x’2

and is read, “the limit of f (x), as x goes to 2 from the right, equals 4.”

35

36 CHAPTER 2. LIMITS AND CONTINUITY

The statement that f (x) tends to 4 as x tends to 2 from the left is written

lim f (x) = 4

x’2’

and is read, “the limit of f (x), as x goes to 2 from the left, equals 4.”

The statement that f (x) tends to 4 as x tends to 2 either from the right

or from the left, is written

lim f (x) = 4

x’2

and is read, “the limit of f (x), as x goes to 2, equals 4.”

The statement that f (x) is continuous at x = 2 is expressed by the

equation

lim f (x) = f (2).

x’2

Example 2.1.2 Consider the unit step function as x tends to 0.

0 if x < 0

u(x) =

1 if x ≥ 0.

graph

The function, u(x) tends to 1 as x tends to 0 from the right side. So, we

write

lim u(x) = 1 = u(0).

+

x’0

The limit of u(x) as x tends to 0 from the left equals 0. Hence,

lim u(x) = 0 = u(0).

x’0’

Since

lim u(x) = u(0),

x’0+

we say that u(x) is continuous at 0 from the right. Since

lim u(x) = u(0),

x’0’

2.1. INTUITIVE TREATMENT AND DEFINITIONS 37

we say that u(x) is not continuous at 0 from the left. In this case the jump

at 0 is 1 and is de¬ned by

jump (u(x), 0) = lim u(x) ’ lim u(x)

’

+

x’0 x’0

= 1.

Observe that the graph of u(x) has two pieces that are not joined together.

Every horizontal line with equation y = c, 0 < c < 1, separates the two

pieces of the graph without intersecting the graph of u(x). This kind of

jump discontinuity at a point is called “¬nite jump” discontinuity.

Example 2.1.3 Consider the signum function, sign(x), de¬ned by

1 if x > 0

x

.

=

sign (x) =

|x| ’1 if x < 0

If x > 0, then sign(x) = 1. If x < 0, then sign(x) = ’1. In this case,

lim sign(x) = 1

x’0+

lim sign(x) = ’1

x’0’

jump (sign(x), 0) = 2.

Since sign(x) is not de¬ned at x = 0, it is not continuous at 0.

sin θ

Example 2.1.4 Consider f (θ) = as θ tends to 0.

θ

graph

The point C(cos θ, sin θ) on the unit circle de¬nes sin θ as the vertical

length BC. The radian measure of the angle θ is the arc length DC. It is

38 CHAPTER 2. LIMITS AND CONTINUITY

clear that the vertical length BC and arc length DC get closer to each other

as θ tends to 0 from above. Thus,

graph

sin θ

lim = 1.

θ +

θ’0

For negative θ, sin θ and θ are both negative.

’ sin θ

sin(’θ)

lim = lim = 1.

’θ ’θ

+ θ’0+

θ’0

Hence,

sin θ

lim = 1.

θ

θ’0

This limit can be veri¬ed by numerical computation for small θ.

1

as x tends to 0 and as x tends to ±∞.

Example 2.1.5 Consider f (x) =

x

graph

It is intuitively clear that

1

lim = +∞

x

+

x’0

1

lim =0

x’+∞ x

1

= ’∞

lim

x’0’ x

1

lim = 0.

x’’∞ x

2.1. INTUITIVE TREATMENT AND DEFINITIONS 39

The function f is not continuous at x = 0 because it is not de¬ned for x = 0.

This discontinuity is not removable because the limits from the left and from

the right, at x = 0, are not equal. The horizontal and vertical axes divide

the graph of f in two separate pieces. The vertical axis is called the vertical

asymptote of the graph of f . The horizontal axis is called the horizontal

asymptote of the graph of f . We say that f has an essential discontinuity at

x = 0.

Example 2.1.6 Consider f (x) = sin(1/x) as x tends to 0.

graph

The period of the sine function is 2π. As observed in Example 5, 1/x

becomes very large as x becomes small. For this reason, many cycles of the

sine wave pass from the value ’1 to the value +1 and a rapid oscillation

occurs near zero. None of the following limits exist:

1 1

1

, lim sin , lim sin .

lim sin

x x x

’ x’0

+ x’0

x’0

It is not possible to de¬ne the function f at 0 to make it continuous. This

kind of discontinuity is called an “oscillation” type of discontinuity.

1

as x tends to 0.

Example 2.1.7 Consider f (x) = x sin

x

graph

1

In this example, sin , oscillates as in Example 6, but the amplitude

x

40 CHAPTER 2. LIMITS AND CONTINUITY

|x| tends to zero as x tends to 0. In this case,

1

lim x sin =0

x

+

x’0

1

lim x sin =0

x

’

x’0

1

lim x sin = 0.

x

x’0

The discontinuity at x = 0 is removable. We de¬ne f (0) = 0 to make f

continuous at x = 0.

x’2

as x tends to ±2.

Example 2.1.8 Consider f (x) =

x2 ’ 4

This is an example of a rational function that yields the indeterminate

form 0/0 when x is replaced by 2. When this kind of situation occurs in

rational functions, it is necessary to cancel the common factors of the nu-

merator and the denominator to determine the appropriate limit if it exists.

In this example, x ’ 2 is the common factor and the reduced form is obtained

through cancellation.

graph

x’2

x’2

=

f (x) =

x2 ’ 4 (x ’ 2)(x + 2)

1

= .

x+2

In order to get the limits as x tends to 2, we used the reduced form to get

1/4. The discontinuity at x = 2 is removed if we de¬ne f (2) = 1/4. This

function still has the essential discontinuity at x = ’2.

2.1. INTUITIVE TREATMENT AND DEFINITIONS 41

√

√

x’ 3

Example 2.1.9 Consider f (x) = as x tends to 3.

x2 ’ 9

In this case f is not a rational function; still, the problem at x = 3 is

√

√

caused by the common factor ( x ’ 3).

graph

√

√

x’ 3

f (x) =

x2 ’ 9 √

√

( x ’ 3)

√√ √

= √

(x + 3)( x ’ 3)( x + 3)

1

√.

= √

(x + 3)( x + 3)

√

As x tends to 3, the reduced form of f tends to 1/(12 3). Thus,

1

√.

lim f (x) = lim f (x) = lim f (x) =

12 3

’ x’3

+

x’3 x’3

1

The discontinuity of f at x = 3 is removed by de¬ning f (3) = √ . The

12 3

√

other discontinuities of f at x = ’3 and x = ’ 3 are essential discontinuities

and cannot be removed.

Even though calculus began intuitively, formal and precise de¬nitions of

limit and continuity became necessary. These precise de¬nitions have become

the foundations of calculus and its applications to the sciences. Let us assume

that a function f is de¬ned in some open interval, (a, b), except possibly at

one point c, such that a < c < b. Then we make the following de¬nitions

using the Greek symbols: , read “epsilon” and δ, read, “delta.”

2.1.2 Limit: Formal De¬nitions

42 CHAPTER 2. LIMITS AND CONTINUITY

De¬nition 2.1.1 The limit of f (x) as x goes to c from the right is L, if and

only if, for each > 0, there exists some δ > 0 such that

|f (x) ’ L| < , whenever, c < x < c + δ.

The statement that the limit of f (x) as x goes to c from the right is L, is

expressed by the equation

lim f (x) = L.

+

x’c

graph

De¬nition 2.1.2 The limit of f (x) as x goes to c from the left is L, if and

only if, for each > 0, there exists some δ > 0 such that

|f (x) ’ L| < , whenever, c ’ δ < x < c.

The statement that the limit of f (x) as x goes to c from the left is L, is

written as

lim f (x) = L.

’

x’c

graph

De¬nition 2.1.3 The (two-sided) limit of f (x) as x goes to c is L, if and

only if, for each > 0, there exists some δ > 0 such that

|f (x) ’ L| < , whenever 0 < |x ’ c| < δ.

graph

2.1. INTUITIVE TREATMENT AND DEFINITIONS 43

The equation

lim f (x) = L

x’c

is read “the (two-sided) limit of f (x) as x goes to c equals L.”

2.1.3 Continuity: Formal De¬nitions

De¬nition 2.1.4 The function f is said to be continuous at c from the right

if f (c) is de¬ned, and

lim f (x) = f (c).

+ x’c

De¬nition 2.1.5 The function f is said to be continuous at c from the left

if f (c) is de¬ned, and

lim f (x) = f (c).

’ x’c

De¬nition 2.1.6 The function f is said to be (two-sided) continuous at c if

f (c) is de¬ned, and

lim f (x) = f (c).

x’c

Remark 4 The continuity de¬nition requires that the following conditions

be met if f is to be continuous at c:

(i) f (c) is de¬ned as a ¬nite real number,

(ii) lim f (x) exists and equals f (c),

’

x’c

(iii) lim f (x) exists and equals f (c),

+

x’c

(iv) lim f (x) = f (c) = lim f (x).

’ +

x’c x’c

When a function f is not continuous at c, one, or more, of these conditions

are not met.

44 CHAPTER 2. LIMITS AND CONTINUITY

Remark 5 All polynomials, sin x, cos x, ex , sinh x, cosh x, bx , b = 1 are con-

tinuous for all real values of x. All logarithmic functions, logb x, b > 0, b = 1

are continuous for all x > 0. Each rational function, p(x)/q(x), is continuous

where q(x) = 0. Each of the functions tan x, cot x, sec x, csc x, tanh x, coth x,

sech x, and csch x is continuous at each point of its domain.

De¬nition 2.1.7 (Algebra of functions) Let f and g be two functions that

have a common domain, say D. Then we de¬ne the following for all x in D:

1. (f + g)(x) = f (x) + g(x) (sum of f and g)

2. (f ’ g)(x) = f (x) ’ g(x) (di¬erence of f and g)

f (x)

f

(x) = , if g(x) = 0 (quotient of f and g)

3.

g g(x)

4. (gf )(x) = g(x)f (x) (product of f and g)

If the range of f is a subset of the domain of g, then we de¬ne the

composition, g —¦ f , of f followed by g, as follows:

5. (g —¦ f )(x) = g(f (x))

Remark 6 The following theorems on limits and continuity follow from the

de¬nitions of limit and continuity.

Theorem 2.1.1 Suppose that for some real numbers L and M , lim f (x) = L

x’c

and lim g(x) = M . Then

x’c

(i) lim k = k, where k is a constant function.

x’c

(ii) lim (f (x) + g(x)) = lim f (x) + lim g(x)

x’c x’c x’c

(iii) lim (f (x) ’ g(x)) = lim f (x) ’ lim g(x)

x’c x’c x’c

2.1. INTUITIVE TREATMENT AND DEFINITIONS 45

(iv) lim (f (x)g(x)) = lim f (x) lim g(x)

x’c x’c x’c

lim f (x)

f (x) x’c

= , if lim g(x) = 0

(v) lim

g(x) lim g(x) x’c

x’c

x’c

Proof.

Part (i) Let f (x) = k for all x and > 0 be given. Then

|f (x) ’ k| = |k ’ k| = 0 <

for all x. This completes the proof of Part (i).

For Parts (ii)“(v) let > 0 be given and let

lim f (x) = L and lim g(x) = M.

x’c x’c

By de¬nition there exist δ1 > 0 and δ2 > 0 such that

|f (x) ’ L| < whenever 0 < |x ’ c| < δ1 (1)

3

|g(x) ’ M | < whenever 0 < |x ’ c| < δ2 (2)

3

Part (ii) Let δ = min(δ1 , δ2 ). Then 0 < |x ’ c| < δ implies that

0 < |x ’ c| < δ1 and |f (x) ’ L| < (by (1)) (3)

3

0 < |x ’ c| < δ2 and |g(x) ’ M | < (by (2)) (4)

3

Hence, if 0 < |x ’ c| < δ, then

|(f (x) + g(x)) ’ (L + M )| = |(f (x) ’ L) + (g(x) ’ M )|

¤ |f (x) ’ L| + |g(x) ’ M |

< + (by (3) and (4))

3 3

<.

This completes the proof of Part (ii).

46 CHAPTER 2. LIMITS AND CONTINUITY

Part (iii) Let δ be de¬ned as in Part (ii). Then 0 < |x ’ c| < δ implies that

|(f (x) ’ g(x)) ’ (L ’ M )| = |(f (x) ’ L) + (g(x) ’ M )|

¤ |f (x) ’ L| + |g(x) ’ M |

< +

3 3

<.

This completes the proof of Part (iii).

Part (iv) Let > 0 be given. Let

.

= min 1,

1

1 + |L| + |M |

Then > 0 and, by de¬nition, there exist δ1 and δ2 such that

1

|f (x) ’ L| < whenever 0 < |x ’ c| < δ1 (5)

1

|g(x) ’ M | < whenever 0 < |x ’ c| < δ2 (6)

1

Let δ = min(δ1 , δ2 ). Then 0 < |x ’ c| < δ implies that

0 < |x ’ c| < δ1 and |f (x) ’ L| < (by (5)) (7)

1

0 < |x ’ c| < δ2 and |g(x) ’ M | < (by (6)) (8)

1

Also,

|f (x)g(x) ’ LM | = |(f (x) ’ L + L)(g(x) ’ M + M ) ’ LM |

= |(f (x) ’ L)(g(x) ’ M ) + (f (x) ’ L)M + L(g(x) ’ M )|

¤ |f (x) ’ L| |g(x) ’ M | + |f (x) + L| |M | + |L| |g(x) ’ M |

< 2 + |M | 1 + |L| 1

1

¤ 1 + |M | 1 + |L| 1

= (1 + |M | + |N |) 1

¤.

This completes the proof of Part (iv).

Part (v) Suppose that M > 0 and lim g(x) = M . Then we show that

x’c

1 1

lim = .

g(x) M

x’c

2.1. INTUITIVE TREATMENT AND DEFINITIONS 47

Since M/2 > 0, there exists some δ1 > 0 such that

M

|g(x) ’ M | < whenever 0 < |x ’ c| < δ1 ,

2

M 3M

’ + M < g(x) < whenever 0 < |x ’ c| < δ1 ,

2 2

M 3M

whenever 0 < |x ’ c| < δ1 ,

0< < g(x) <

2 2

1 2

whenever 0 < |x ’ c| < δ1 .

<

|g(x)| M

Let > 0 be given. Let 1 = M 2 /2. Then > 0 and there exists some

1

δ > 0 such that δ < δ1 and

|g(x) ’ M | < 1 whenever 0 < |x ’ c| < δ < δ1 ,

M ’ g(x) |g(x) ’ M |

1 1

’ = =

|g(x)|M

g(x) M g(x)M

1 1

· |g(x) ’ M |

=

M |g(x)|

12

· ·1

<

MM

21

=2

M

whenever 0 < |x ’ c| < δ.

=

This completes the proof of the statement

1 1

lim = whenever M > 0.

g(x) M

x’c

The case for M < 0 can be proven in a similar manner. Now, we can use

Part (iv) to prove Part (v) as follows:

1

f (x)

= lim f (x) ·

lim

x’c g(x) g(x)

x’c

1

= lim f (x) · lim

g(x)

x’c x’c

1

=L·

M

L

= .

M

48 CHAPTER 2. LIMITS AND CONTINUITY

This completes the proof of Theorem 2.1.1.

Theorem 2.1.2 If f and g are two functions that are continuous on a com-

mon domain D, then the sum, f + g, the di¬erence, f ’ g and the product,

f g, are continuous on D. Also, f /g is continuous at each point x in D such

that g(x) = 0.

Proof. If f and g are continuous at c, then f (c) and g(c) are real numbers

and

lim f (x) = f (c), lim g(x) = g(c).

x’c x’c

By Theorem 2.1.1, we get

lim(f (x) + g(x)) = lim f (x) + lim g(x) = f (c) + g(c)

x’c x’c x’c

lim(f (x) ’ g(x)) = lim f (x) ’ lim g(x) = f (c) ’ g(c)

x’c x’c x’c

lim(f (x)g(x)) = lim f (x) lim(g(x)) = f (c)g(c)

x’c x’c x’c

limx’c f (x) f (c)

f (x)

= = , if g(c) = 0.

lim

g(x) limx’c g(x) g(c)

x’c

This completes the proof of Theorem 2.1.2.

2.1.4 Continuity Examples

Example 2.1.10 Show that the constant function f (x) = 4 is continuous at

every real number c. Show that for every constant k, f (x) = k is continuous

at every real number c.

First of all, if f (x) = 4, then f (c) = 4. We need to show that

lim 4 = 4.

x’c

graph

For each > 0, let δ = 1. Then

|f (x) ’ f (c)| = |4 ’ 4| = 0 <

2.1. INTUITIVE TREATMENT AND DEFINITIONS 49

for all x such that |x ’ c| < 1. Secondly, for each > 0, let δ = 1. Then

|f (x) ’ f (c)| = |k ’ k| = 0 <

for all x such that |x ’ c| < 1. This completes the required proof.

Example 2.1.11 Show that f (x) = 3x ’ 4 is continuous at x = 3.

Let > 0 be given. Then

|f (x) ’ f (3)| = |(3x ’ 4) ’ (5)|

= |3x ’ 9|

= 3|x ’ 3|

<

whenever |x ’ 3| < .

3

We de¬ne δ = . Then, it follows that

3

lim f (x) = f (3)

x’3

and, hence, f is continuous at x = 3.

Example 2.1.12 Show that f (x) = x3 is continuous at x = 2.

Since f (2) = 8, we need to prove that

lim x3 = 8 = 23 .

x’2

graph

Let > 0 be given. Let us concentrate our attention on the open interval

50 CHAPTER 2. LIMITS AND CONTINUITY

(1, 3) that contains x = 2 at its mid-point. Then

|f (x) ’ f (2)| = |x3 ’ 8| = |(x ’ 2)(x2 + 2x + 4)|

= |x ’ 2| |x2 + 2x + 4|

¤ |x ’ 2|(|x|2 + 2|x| + 4) (Triangle Inequality |u + v| ¤ |u| + |v|)

¤ |x ’ 2|(9 + 18 + 4)

= 31|x ’ 2|

<

Provided

|x ’ 2| <

.

31

Since we are concentrating on the interval (1, 3) for which |x ’ 2| < 1, we

need to de¬ne δ to be the minimum of 1 and . Thus, if we de¬ne δ =

31

min{1, /31}, then

|f (x) ’ f (2)| <

whenever |x ’ 2| < δ. By de¬nition, f (x) is continuous at x = 2.

Example 2.1.13 Show that every polynomial P (x) is continuous at every

c.

From algebra, we recall that, by the Remainder Theorem,

P (x) = (x ’ c)Q(x) + P (c).

Thus,

|P (x) ’ P (c)| = |x ’ c||Q(x)|

where Q(x) is a polynomial of degree one less than the degree of P (x). As

in Example 12, |Q(x)| is bounded on the closed interval [c ’ 1, c + 1]. For

example, if

Q(x) = q0 xn’1 + q1 xn’2 + · · · + qn’2 x + qn’1

|Q(x)| ¤ |q0 | |x|n’1 + |q1 | |x|n’2 + · · · + |qn’2 | |x| + |qn’1 |.

Let m = max{|x| : c ’ 1 ¤ x ¤ c + 1}. Then

|Q(x)| ¤ |q0 |mn’1 + |q1 |mn’2 + · · · + qn’2 m + |qn’1 | = M,

2.1. INTUITIVE TREATMENT AND DEFINITIONS 51

for some M . Then

|P (x) ’ P (c)| = |x ’ c| |Q(x)| ¤ M |x ’ c| <

whenever |x ’ c| < . As in Example 12, we de¬ne δ = min 1, . Then

M M

|P (x) ’ P (c)| < , whenever |x ’ c| < δ. Hence,

lim P (x) = P (c)

x’c

and by de¬nition P (x) is continuous at each number c.

1

Example 2.1.14 Show that f (x) = is continuous at every real number

x

c > 0.

We need to show that

1 1

lim =.

x’c x c

c

Let > 0 be given. Let us concentrate on the interval |x ’ c| ¤ ; that is,

2

c 3c

¤ x ¤ . Clearly, x = 0 in this interval. Then

2 2

11

|f (x) ’ f (c)| = ’

xc

c’x

=

cx

11

= |x ’ c| · ·

c |x|

12

< |x ’ c| · ·

cc

2

= 2 |x ’ c|

c

<

c2

whenever |x ’ c| < .

2

c c2

. Then for all x such that |x ’ c| < δ,

We de¬ne δ = min ,

22

11

’ <.

xc

52 CHAPTER 2. LIMITS AND CONTINUITY

Hence,

1 1

lim =

x c

x’c

1

and the function f (x) = is continuous at each c > 0.

x

1

A similar argument can be used for c < 0. The function f (x) = is

x

continuous for all x = 0.

Example 2.1.15 Suppose that the domain of a function g contains an open

interval containing c, and the range of g contains an open interval containing

g(c). Suppose further that the domain of f contains the range of g. Show

that if g is continuous at c and f is continuous at g(c), then the composition

f —¦ g is continuous at c.

We need to show that

lim f (g(x)) = f (g(c)).

x’c

Let > 0 be given. Since f is continuous at g(c), there exists δ1 > 0 such

that

1. |f (y) ’ f (g(c))| < , whenever, |y ’ g(c)| < δ1 .

Since g is continuous at c, and δ1 > 0, there exists δ > 0 such that

2. |g(x) ’ g(c)| < δ1 , whenever, |x ’ c| < δ.

On replacing y by g(x) in equation (1), we get

|f (g(x)) ’ f (g(c))| < , whenever, |x ’ c| < δ.

By de¬nition, it follows that

lim f (g(x)) = f (g(c))

x’c

and the composition f —¦ g is continuous at c.

Example 2.1.16 Suppose that two functions f and g have a common do-

main that contains one open interval containing c. Suppose further that f

and g are continuous at c. Then show that

2.1. INTUITIVE TREATMENT AND DEFINITIONS 53

(i) f + g is continuous at c,

(ii) f ’ g is continuous at c,

(iii) kf is continuous at c for every constant k = 0,

(iv) f · g is continuous at c.

Part (i) We need to prove that

lim [f (x) + g(x)] = f (c) + g(c).

x’c

Let > 0 be given. Then > 0. Since f is continuous at c and > 0, there

2 2

exists some δ1 > 0 such that

|f (x) ’ f (c)| < , whenever, |x ’ c| ¤ δ1 .

(1)

2

Also, since g is continuous at c and > 0, there exists some δ2 > 0 such that

2

δ

|g(x) ’ g(c)| < , whenever, |x ’ c| < .

(2)

2 2

Let δ = min{δ1 , δ2 }. Then δ > 0. Let |x ’ c| < δ. Then |x ’ c| < δ1 and

|x ’ c| < δ2 . For this choice of x, we get

|{f (x) + g(x)} ’ {f (c) + g(c)}|

= |{f (x) ’ f (c)} + {g(x) ’ g(c)}|

¤ |f (x) ’ f (c)| + |g(x) ’ g(c)| (by triangle inequality)

< +

2 2

=.

It follows that

lim (f (x) + g(x)) = f (c) + g(c)

x’0

and f + g is continuous at c. This proves part (i).

54 CHAPTER 2. LIMITS AND CONTINUITY

Part (ii) For Part (ii) we chose , /2, δ1 , δ2 and δ exactly as in Part (i).

Suppose |x ’ c| < δ. Then |x ’ c| < δ1 and |x ’ c| < δ2 . For these choices of

x we get

|{f (x) ’ g(x)} ’ {f (c) ’ g(c)}|

= |{f (x) ’ f (c)} ’ {g(x) ’ g(c)}|

¤ |f (x) ’ f (c)| + |g(x) ’ g(c)| (by triangle inequality)

< +

2 2

=.

It follows that

lim (f (x) ’ g(x)) = f (c) ’ g(c)

x’c

and, hence, f ’ g is continuous at c.

Part (iii) For Part (iii) let > 0 be given. Since k = 0, > 0. Since f is

|k|

continuous at c, there exists some δ > 0 such that

whenever, |x ’ c| < δ.

|f (x) ’ f (c)| < ,

|k|

If |x ’ c| < δ, then

|kf (x) ’ kf (c)| = |k(f (x) ’ f (c))|

= |k| |(f (x) ’ f (c)|

< |k| ·

|k|

=.

It follows that

lim kf (x) = kf (c)

x’c

and, hence, kf is continuous at c.

Part (iv) We need to show that

lim (f (x)g(x)) = f (c)g(c).

x’c

2.1. INTUITIVE TREATMENT AND DEFINITIONS 55

Let > 0 be given. Without loss of generality we may assume that < 1.

. Then 1 > 0, 1 < 1 and 1 (1 + |f | + |g(c)|) =

Let 1=

2(1 + |f (c)| + |g(c)|)

< . Since f is continuous at c and 1 > 0, there exists δ1 > 0 such that

2

|f (x) ’ f (c)| < whenever, |x ’ c| < δ1 .

1

Also, since g is continuous at c and > 0, there exists δ2 > 0 such that

1

|g(x) ’ g(c)| < whenever, |x ’ c| < δ2 .

1

Let δ = min{δ1 , δ2 } and |x ’ c| < δ. For these choices of x, we get

|f (x)g(x) ’ f (c)g(c)|

= |(f (x) ’ f (c) + f (c))(g(x) ’ g(c) + g(c)) ’ f (c)g(c)|

= |(f (x) ’ f (c))(g(x) ’ g(c)) + (f (x) ’ f (c))g(c) + f (c)(g(x) ’ g(c))|

¤ |f (x) ’ f (c)| |g(x) ’ g(c)| + |f (x) ’ f (c)| |g(c)| + |f (c)| |g(x) ’ g(c)|

< 1 · 1 + 1 |g(c)| + 1 |f (c)|

< 1 (1 + |g(c)| + |f (c)|) , (since 1 < 1)

<.

It follows that

lim f (x)g(x) = f (c)g(c)

x’c

and, hence, the product f · g is continuous at c.

Example 2.1.17 Show that the quotient f /g is continuous at c if f and g

are continuous at c and g(c) = 0.

First of all, let us observe that the function 1/g is a composition of g(x)

and 1/x and hence 1/g is continuous at c by virtue of the arguments in

Examples 14 and 15. By the argument in Example 16, the product f (1/g) =

f /g is continuous at c, as required in Example 17.

Example 2.1.18 Show that a rational function of the form p(x)/q(x) is

continuous for all c such that g(c) = 0.

56 CHAPTER 2. LIMITS AND CONTINUITY

In Example 13, we showed that each polynomial function is continuous

at every real number c. Therefore, p(x) is continuous at every c and q(x) is

continuous at every c. By virtue of the argument in Example 17, the quotient

p(x)/q(x) is continuous for all c such that q(c) = 0.

Example 2.1.19 Suppose that f (x) ¤ g(x) ¤ h(x) for all x in an open

interval containing c and

lim f (x) = lim h(x) = L.

x’c x’c

Then, show that,

lim g(x) = L.

x’c

Let > 0 be given. Then there exist δ1 > 0, δ2 > 0, and δ = min{δ1 , δ2 }

such that

whenever 0 < |x ’ c| < δ1

|f (x) ’ L| <

2

|h(x) ’ L) < whenever 0 < |x ’ c| < δ2 .

2

If 0 < |x ’ c| < δ1 , then 0 < |x ’ c| < δ1 , 0 < |x ’ c| < δ2 and, hence,

’ < f (x) ’ L < g(x) ’ L < h(x) ’ L < .

2 2

It follows that

|g(x) ’ L| < < whenever 0 < |x ’ c| < δ,

2

and

lim g(x) = L.

x’c

Example 2.1.20 Show that f (x) = |x| is continuous at 0.

We need to show that

lim |x| = 0.

x’0

Let > 0 be given. Let δ = . Then |x ’ 0| < implies that |x| < Hence,

lim |x| = 0

x’0

2.1. INTUITIVE TREATMENT AND DEFINITIONS 57

Example 2.1.21 Show that

(i) lim sin θ = 0 (ii) lim cos θ = 1

θ’0 θ’0

1 ’ cos θ

sin θ

(iii) lim =1 (iv) lim =0

θ θ

θ’0 θ’0

graph

Part (i) By de¬nition, the point C(cos θ, sin θ), where θ is the length of

the arc CD, lies on the unit circle. It is clear that the length BC = sin θ is

less than θ, the arclength of the arc CD, for small positive θ. Hence,

’θ ¤ sin θ ¤ θ

and

lim sin θ = 0.

θ’0+

For small negative θ, we get

θ ¤ sin θ ¤ ’θ

and

lim sin θ = 0.

θ’0’

Therefore,

lim sin θ = 0.

θ’0

Part (ii) It is clear that the point B approaches D as θ tends to zero. There-

fore,

lim cos θ = 1.

θ’0

Part (iii) Consider the inequality

Area of triangle ABC ¤ Area of sector ADC ¤ Area of triangle ADE

1 1 sin θ

1

cos θ sin θ ¤ θ ¤ .

2 2 2 cos θ

58 CHAPTER 2. LIMITS AND CONTINUITY

Assume that θ is small but positive. Multiply each part of the inequality by

2/ sin θ to get

θ 1

cos θ ¤ ¤ .

sin θ cos θ

On taking limits and using the squeeze theorem, we get

θ

lim = 1.

sin θ

+

θ’0

By taking reciprocals, we get