<<

. 2
( 14)



>>

=.
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

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