<< ŮÚū. 2(‚ŮŚ„Ó 14)—őńŇ–∆ņÕ»Ň >>
=.
1 + ex 3
On multiplying through, we get
1
3ex = 1 + ex or 2ex = 1, ex =
2
x = ln(1/2) = ‚ą’ ln(2).

Example 1.4.4 Prove that for all real x, cosh2 x ‚ą’ sinh2 x = 1.
2 2
1 1
cosh2 x ‚ą’ sinh2 x = (ex + e‚ą’x ) ‚ą’ (ex ‚ą’ e‚ą’x )
2 2
1
= [e2x + 2 + e‚ą’2x ) ‚ą’ (e2x ‚ą’ 2 + e‚ą’2x )]
4
1
= 
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 ‚ą’ e‚ą’x ) ¬· (ey + e‚ą’y )
2 2
1 1
+ (ex + e‚ą’x ) ¬· (ey ‚ą’ e‚ą’y )
2 2
1
= [(ex+y + ex‚ą’y ‚ą’ e‚ą’x+y ‚ą’ e‚ą’x‚ą’y )
4
+ (ex+y ‚ą’ ex‚ą’y + e‚ą’x+y ‚ą’ e‚ą’x‚ą’y )]
1
= [2(ex+y ‚ą’ e‚ą’(x+y) ]
4
1
= (e(x+y) ‚ą’ e‚ą’(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 ‚ą’ e‚ą’x ). Then
2
1x
(e ‚ą’ e‚ą’x )
2ex y = 2ex = e2x ‚ą’ 1
2
e2x ‚ą’ 2yex ‚ą’ 1 = 0
(ex )2 ‚ą’ (2y)ex ‚ą’ 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 deÔ¬Āned by
‚ąö
arcsinh x = ln(x + 1 + x2 )

(b) As in part (a), we let y = cosh x and
1
2ex y = 2ex ¬· (ex + e‚ą’x ) = e2x + 1
2
2x x
e ‚ą’ (2y)e + 1 = 0
4y 2 ‚ą’ 4
2y ¬±
x
e=
2
ex = y ¬± y 2 ‚ą’ 1.

We observe that cosh x is an even function and hence it is not one-to-
one. Since cosh(‚ą’x) = cosh(x), we will solve for the larger x. On taking
natural logarithms of both sides, we get

y 2 ‚ą’ 1) or x2 = ln(y ‚ą’ y 2 ‚ą’ 1).
x1 = ln(y +

We observe that

y 2 ‚ą’ 1)(y + y 2 ‚ą’ 1)
(y ‚ą’
y2
x2 = ln(y ‚ą’ ‚ą’ 1) = ln
y2 ‚ą’ 1
y+
1
= ln
y2 ‚ą’ 1
y+
y 2 ‚ą’ 1) = ‚ą’x1 .
= ‚ą’ ln(y +

Thus, we can deÔ¬Āne, as the principal branch,
‚ąö
x2 ‚ą’ 1), x ‚Č• 1
arccosh x = ln(x +
32 CHAPTER 1. FUNCTIONS

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

ex ‚ą’ e‚ą’x
|y| < 1
y= x ,
e + e‚ą’x
ex [(ex + e‚ą’x )y] = ex [(ex ‚ą’ e‚ą’x )] |y| < 1
,
(e2x + 1)y = e2x ‚ą’ 1 |y| < 1
,
e2x (y ‚ą’ 1) = ‚ą’(1 + y) |y| < 1
,
(1 + y)
e2x = ‚ą’ |y| < 1
,
y‚ą’1
1+y
e2x = |y| < 1
,
1‚ą’y
1+y
|y| < 1
2x = ln ,
1‚ą’y
1 1+y
, |y| < 1.
x = ln
1‚ą’y
2

Therefore, the inverse of the function tanh x, denoted arctanhx, is deÔ¬Āned
by
1 1+x
, |x| < 1.
arctanh , x = ln
1‚ą’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 ‚ą’2 )
(e) eln(e
(d) log10
(.0001)2/3

2. Prove each of the following identities

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

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

3. Simplify the radical expression by using the given substitution.
‚ąö ‚ąö
2 + x2 , x = a sinh t (b) x2 ‚ą’ a2 , x = a cosh t
(a) a
‚ąö
(c) a2 ‚ą’ 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 = , Ô¬Ānd 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‚ą’x d) y = 2‚ą’x
a) y = 10x b) y = 2x
2 2
f) y = e‚ą’x g) y = xe‚ą’x i) y = e‚ą’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 ‚ą’ ln(x + 1) = ln(4) b) 2 log10 (x ‚ą’ 3) = log10 (x + 5) + log10 4

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

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

Limits and Continuity

2.1 Intuitive treatment and deÔ¬Ānitions
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‚Ü’2

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

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‚Ü’2‚ą’

and is read, ‚Äúthe limit of f (x), as x goes to 2 from the left, equals 4.‚ÄĚ
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‚Ü’2

and is read, ‚Äúthe limit of f (x), as x goes to 2, equals 4.‚ÄĚ
The statement that f (x) is continuous at x = 2 is expressed by the
equation
lim f (x) = f (2).
x‚Ü’2

Example 2.1.2 Consider the unit step function as x tends to 0.
0 if x < 0
u(x) =
1 if x ‚Č• 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‚Ü’0

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

lim u(x) = 0 = u(0).
x‚Ü’0‚ą’

Since
lim u(x) = u(0),
x‚Ü’0+

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

lim u(x) = u(0),
x‚Ü’0‚ą’
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 deÔ¬Āned by

jump (u(x), 0) = lim u(x) ‚ą’ lim u(x)
‚ą’
+
x‚Ü’0 x‚Ü’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 ‚ÄúÔ¬Ānite jump‚ÄĚ discontinuity.

Example 2.1.3 Consider the signum function, sign(x), deÔ¬Āned by

1 if x > 0
x
.
=
sign (x) =
|x| ‚ą’1 if x < 0

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

lim sign(x) = 1
x‚Ü’0+
lim sign(x) = ‚ą’1
x‚Ü’0‚ą’
jump (sign(x), 0) = 2.

Since sign(x) is not deÔ¬Āned at x = 0, it is not continuous at 0.

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

graph

The point C(cos őł, sin őł) on the unit circle deÔ¬Ānes sin őł as the vertical
length BC. The radian measure of the angle őł 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 őł tends to 0 from above. Thus,

graph

sin őł
lim = 1.
őł +
őł‚Ü’0
For negative őł, sin őł and őł are both negative.
‚ą’ sin őł
sin(‚ą’őł)
lim = lim = 1.
‚ą’őł ‚ą’őł
+ őł‚Ü’0+
őł‚Ü’0

Hence,
sin őł
lim = 1.
őł
őł‚Ü’0
This limit can be veriÔ¬Āed by numerical computation for small őł.

1
as x tends to 0 and as x tends to ¬±‚ąě.
Example 2.1.5 Consider f (x) =
x

graph

It is intuitively clear that
1
lim = +‚ąě
x
+
x‚Ü’0
1
lim =0
x‚Ü’+‚ąě x
1
= ‚ą’‚ąě
lim
x‚Ü’0‚ą’ x
1
lim = 0.
x‚Ü’‚ą’‚ąě x
2.1. INTUITIVE TREATMENT AND DEFINITIONS 39

The function f is not continuous at x = 0 because it is not deÔ¬Āned 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 2ŌÄ. 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 ‚ą’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
‚ą’ x‚Ü’0
+ x‚Ü’0
x‚Ü’0

It is not possible to deÔ¬Āne the function f at 0 to make it continuous. This
kind of discontinuity is called an ‚Äúoscillation‚ÄĚ 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‚Ü’0

1
lim x sin =0
x
‚ą’
x‚Ü’0

1
lim x sin = 0.
x
x‚Ü’0

The discontinuity at x = 0 is removable. We deÔ¬Āne f (0) = 0 to make f
continuous at x = 0.

x‚ą’2
as x tends to ¬±2.
Example 2.1.8 Consider f (x) =
x2 ‚ą’ 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 ‚ą’ 2 is the common factor and the reduced form is obtained
through cancellation.

graph

x‚ą’2
x‚ą’2
=
f (x) =
x2 ‚ą’ 4 (x ‚ą’ 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 deÔ¬Āne f (2) = 1/4. This
function still has the essential discontinuity at x = ‚ą’2.
2.1. INTUITIVE TREATMENT AND DEFINITIONS 41
‚ąö
‚ąö
x‚ą’ 3
Example 2.1.9 Consider f (x) = as x tends to 3.
x2 ‚ą’ 9
In this case f is not a rational function; still, the problem at x = 3 is
‚ąö
‚ąö
caused by the common factor ( x ‚ą’ 3).

graph

‚ąö
‚ąö
x‚ą’ 3
f (x) =
x2 ‚ą’ 9 ‚ąö
‚ąö
( x ‚ą’ 3)
‚ąö‚ąö ‚ąö
= ‚ąö
(x + 3)( x ‚ą’ 3)( x + 3)
1
‚ąö.
= ‚ąö
(x + 3)( x + 3)
‚ąö
As x tends to 3, the reduced form of f tends to 1/(12 3). Thus,

1
‚ąö.
lim f (x) = lim f (x) = lim f (x) =
12 3
‚ą’ x‚Ü’3
+
x‚Ü’3 x‚Ü’3

1
The discontinuity of f at x = 3 is removed by deÔ¬Āning f (3) = ‚ąö . The
12 3
‚ąö
other discontinuities of f at x = ‚ą’3 and x = ‚ą’ 3 are essential discontinuities
and cannot be removed.
Even though calculus began intuitively, formal and precise deÔ¬Ānitions of
limit and continuity became necessary. These precise deÔ¬Ānitions have become
the foundations of calculus and its applications to the sciences. Let us assume
that a function f is deÔ¬Āned in some open interval, (a, b), except possibly at
one point c, such that a < c < b. Then we make the following deÔ¬Ānitions

2.1.2 Limit: Formal DeÔ¬Ānitions
42 CHAPTER 2. LIMITS AND CONTINUITY

DeÔ¬Ānition 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 őī > 0 such that

|f (x) ‚ą’ L| < , whenever, c < x < c + őī.

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‚Ü’c

graph

DeÔ¬Ānition 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 őī > 0 such that

|f (x) ‚ą’ L| < , whenever, c ‚ą’ őī < 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.
‚ą’
x‚Ü’c

graph

DeÔ¬Ānition 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 őī > 0 such that

|f (x) ‚ą’ L| < , whenever 0 < |x ‚ą’ c| < őī.

graph
2.1. INTUITIVE TREATMENT AND DEFINITIONS 43

The equation
lim f (x) = L
x‚Ü’c

is read ‚Äúthe (two-sided) limit of f (x) as x goes to c equals L.‚ÄĚ

2.1.3 Continuity: Formal DeÔ¬Ānitions
DeÔ¬Ānition 2.1.4 The function f is said to be continuous at c from the right
if f (c) is deÔ¬Āned, and
lim f (x) = f (c).
+ x‚Ü’c

DeÔ¬Ānition 2.1.5 The function f is said to be continuous at c from the left
if f (c) is deÔ¬Āned, and
lim f (x) = f (c).
‚ą’ x‚Ü’c

DeÔ¬Ānition 2.1.6 The function f is said to be (two-sided) continuous at c if
f (c) is deÔ¬Āned, and
lim f (x) = f (c).
x‚Ü’c

Remark 4 The continuity deÔ¬Ānition requires that the following conditions
be met if f is to be continuous at c:

(i) f (c) is deÔ¬Āned as a Ô¬Ānite real number,

(ii) lim f (x) exists and equals f (c),
‚ą’
x‚Ü’c

(iii) lim f (x) exists and equals f (c),
+
x‚Ü’c

(iv) lim f (x) = f (c) = lim f (x).
‚ą’ +
x‚Ü’c x‚Ü’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.

DeÔ¬Ānition 2.1.7 (Algebra of functions) Let f and g be two functions that
have a common domain, say D. Then we deÔ¬Āne the following for all x in D:

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

2. (f ‚ą’ g)(x) = f (x) ‚ą’ g(x) (diÔ¬Äerence 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 deÔ¬Āne 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
deÔ¬Ānitions of limit and continuity.

Theorem 2.1.1 Suppose that for some real numbers L and M , lim f (x) = L
x‚Ü’c
and lim g(x) = M . Then
x‚Ü’c

(i) lim k = k, where k is a constant function.
x‚Ü’c

(ii) lim (f (x) + g(x)) = lim f (x) + lim g(x)
x‚Ü’c x‚Ü’c x‚Ü’c

(iii) lim (f (x) ‚ą’ g(x)) = lim f (x) ‚ą’ lim g(x)
x‚Ü’c x‚Ü’c x‚Ü’c
2.1. INTUITIVE TREATMENT AND DEFINITIONS 45

(iv) lim (f (x)g(x)) = lim f (x) lim g(x)
x‚Ü’c x‚Ü’c x‚Ü’c

lim f (x)
f (x) x‚Ü’c
= , if lim g(x) = 0
(v) lim
g(x) lim g(x) x‚Ü’c
x‚Ü’c
x‚Ü’c

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

|f (x) ‚ą’ k| = |k ‚ą’ k| = 0 <

for all x. This completes the proof of Part (i).
For Parts (ii)‚Ä“(v) let > 0 be given and let

lim f (x) = L and lim g(x) = M.
x‚Ü’c x‚Ü’c

By deÔ¬Ānition there exist őī1 > 0 and őī2 > 0 such that

|f (x) ‚ą’ L| < whenever 0 < |x ‚ą’ c| < őī1 (1)
3
|g(x) ‚ą’ M | < whenever 0 < |x ‚ą’ c| < őī2 (2)
3

Part (ii) Let őī = min(őī1 , őī2 ). Then 0 < |x ‚ą’ c| < őī implies that

0 < |x ‚ą’ c| < őī1 and |f (x) ‚ą’ L| < (by (1)) (3)
3
0 < |x ‚ą’ c| < őī2 and |g(x) ‚ą’ M | < (by (2)) (4)
3
Hence, if 0 < |x ‚ą’ c| < őī, then

|(f (x) + g(x)) ‚ą’ (L + M )| = |(f (x) ‚ą’ L) + (g(x) ‚ą’ M )|
‚Č¤ |f (x) ‚ą’ L| + |g(x) ‚ą’ M |
< + (by (3) and (4))
3 3
<.

This completes the proof of Part (ii).
46 CHAPTER 2. LIMITS AND CONTINUITY

Part (iii) Let őī be deÔ¬Āned as in Part (ii). Then 0 < |x ‚ą’ c| < őī implies that
|(f (x) ‚ą’ g(x)) ‚ą’ (L ‚ą’ M )| = |(f (x) ‚ą’ L) + (g(x) ‚ą’ M )|
‚Č¤ |f (x) ‚ą’ L| + |g(x) ‚ą’ 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 deÔ¬Ānition, there exist őī1 and őī2 such that
1

|f (x) ‚ą’ L| < whenever 0 < |x ‚ą’ c| < őī1 (5)
1
|g(x) ‚ą’ M | < whenever 0 < |x ‚ą’ c| < őī2 (6)
1

Let őī = min(őī1 , őī2 ). Then 0 < |x ‚ą’ c| < őī implies that
0 < |x ‚ą’ c| < őī1 and |f (x) ‚ą’ L| < (by (5)) (7)
1
0 < |x ‚ą’ c| < őī2 and |g(x) ‚ą’ M | < (by (6)) (8)
1

Also,
|f (x)g(x) ‚ą’ LM | = |(f (x) ‚ą’ L + L)(g(x) ‚ą’ M + M ) ‚ą’ LM |
= |(f (x) ‚ą’ L)(g(x) ‚ą’ M ) + (f (x) ‚ą’ L)M + L(g(x) ‚ą’ M )|
‚Č¤ |f (x) ‚ą’ L| |g(x) ‚ą’ M | + |f (x) + L| |M | + |L| |g(x) ‚ą’ M |
< 2 + |M | 1 + |L| 1
1
‚Č¤ 1 + |M | 1 + |L| 1
= (1 + |M | + |N |) 1
‚Č¤.
This completes the proof of Part (iv).

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

1 1
lim = .
g(x) M
x‚Ü’c
2.1. INTUITIVE TREATMENT AND DEFINITIONS 47

Since M/2 > 0, there exists some őī1 > 0 such that
M
|g(x) ‚ą’ M | < whenever 0 < |x ‚ą’ c| < őī1 ,
2
M 3M
‚ą’ + M < g(x) < whenever 0 < |x ‚ą’ c| < őī1 ,
2 2
M 3M
whenever 0 < |x ‚ą’ c| < őī1 ,
0< < g(x) <
2 2
1 2
whenever 0 < |x ‚ą’ c| < őī1 .
<
|g(x)| M
Let > 0 be given. Let 1 = M 2 /2. Then > 0 and there exists some
1
őī > 0 such that őī < őī1 and
|g(x) ‚ą’ M | < 1 whenever 0 < |x ‚ą’ c| < őī < őī1 ,
M ‚ą’ g(x) |g(x) ‚ą’ M |
1 1
‚ą’ = =
|g(x)|M
g(x) M g(x)M
1 1
¬· |g(x) ‚ą’ M |
=
M |g(x)|
12
¬· ¬·1
<
MM
21
=2
M
whenever 0 < |x ‚ą’ c| < őī.
=
This completes the proof of the statement
1 1
lim = whenever M > 0.
g(x) M
x‚Ü’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‚Ü’c g(x) g(x)
x‚Ü’c

1
= lim f (x) ¬· lim
g(x)
x‚Ü’c x‚Ü’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 diÔ¬Äerence, f ‚ą’ 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‚Ü’c x‚Ü’c
By Theorem 2.1.1, we get
lim(f (x) + g(x)) = lim f (x) + lim g(x) = f (c) + g(c)
x‚Ü’c x‚Ü’c x‚Ü’c
lim(f (x) ‚ą’ g(x)) = lim f (x) ‚ą’ lim g(x) = f (c) ‚ą’ g(c)
x‚Ü’c x‚Ü’c x‚Ü’c

lim(f (x)g(x)) = lim f (x) lim(g(x)) = f (c)g(c)
x‚Ü’c x‚Ü’c x‚Ü’c
limx‚Ü’c f (x) f (c)
f (x)
= = , if g(c) = 0.
lim
g(x) limx‚Ü’c g(x) g(c)
x‚Ü’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‚Ü’c

graph

For each > 0, let őī = 1. Then
|f (x) ‚ą’ f (c)| = |4 ‚ą’ 4| = 0 <
2.1. INTUITIVE TREATMENT AND DEFINITIONS 49

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

|f (x) ‚ą’ f (c)| = |k ‚ą’ k| = 0 <

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

Example 2.1.11 Show that f (x) = 3x ‚ą’ 4 is continuous at x = 3.
Let > 0 be given. Then

|f (x) ‚ą’ f (3)| = |(3x ‚ą’ 4) ‚ą’ (5)|
= |3x ‚ą’ 9|
= 3|x ‚ą’ 3|
<

whenever |x ‚ą’ 3| < .
3
We deÔ¬Āne őī = . Then, it follows that
3
lim f (x) = f (3)
x‚Ü’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‚Ü’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) ‚ą’ f (2)| = |x3 ‚ą’ 8| = |(x ‚ą’ 2)(x2 + 2x + 4)|
= |x ‚ą’ 2| |x2 + 2x + 4|
‚Č¤ |x ‚ą’ 2|(|x|2 + 2|x| + 4) (Triangle Inequality |u + v| ‚Č¤ |u| + |v|)
‚Č¤ |x ‚ą’ 2|(9 + 18 + 4)
= 31|x ‚ą’ 2|
<

Provided
|x ‚ą’ 2| <
.
31
Since we are concentrating on the interval (1, 3) for which |x ‚ą’ 2| < 1, we
need to deÔ¬Āne őī to be the minimum of 1 and . Thus, if we deÔ¬Āne őī =
31
min{1, /31}, then
|f (x) ‚ą’ f (2)| <
whenever |x ‚ą’ 2| < őī. By deÔ¬Ānition, 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 ‚ą’ c)Q(x) + P (c).

Thus,
|P (x) ‚ą’ P (c)| = |x ‚ą’ 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 ‚ą’ 1, c + 1]. For
example, if
Q(x) = q0 xn‚ą’1 + q1 xn‚ą’2 + ¬· ¬· ¬· + qn‚ą’2 x + qn‚ą’1
|Q(x)| ‚Č¤ |q0 | |x|n‚ą’1 + |q1 | |x|n‚ą’2 + ¬· ¬· ¬· + |qn‚ą’2 | |x| + |qn‚ą’1 |.
Let m = max{|x| : c ‚ą’ 1 ‚Č¤ x ‚Č¤ c + 1}. Then

|Q(x)| ‚Č¤ |q0 |mn‚ą’1 + |q1 |mn‚ą’2 + ¬· ¬· ¬· + qn‚ą’2 m + |qn‚ą’1 | = M,
2.1. INTUITIVE TREATMENT AND DEFINITIONS 51

for some M . Then
|P (x) ‚ą’ P (c)| = |x ‚ą’ c| |Q(x)| ‚Č¤ M |x ‚ą’ c| <

whenever |x ‚ą’ c| < . As in Example 12, we deÔ¬Āne őī = min 1, . Then
M M
|P (x) ‚ą’ P (c)| < , whenever |x ‚ą’ c| < őī. Hence,
lim P (x) = P (c)
x‚Ü’c

and by deÔ¬Ānition 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‚Ü’c x c
c
Let > 0 be given. Let us concentrate on the interval |x ‚ą’ c| ‚Č¤ ; that is,
2
c 3c
‚Č¤ x ‚Č¤ . Clearly, x = 0 in this interval. Then
2 2
11
|f (x) ‚ą’ f (c)| = ‚ą’
xc
c‚ą’x
=
cx
11
= |x ‚ą’ c| ¬· ¬·
c |x|
12
< |x ‚ą’ c| ¬· ¬·
cc
2
= 2 |x ‚ą’ c|
c
<
c2
whenever |x ‚ą’ c| < .
2
c c2
. Then for all x such that |x ‚ą’ c| < őī,
We deÔ¬Āne őī = min ,
22
11
‚ą’ <.
xc
52 CHAPTER 2. LIMITS AND CONTINUITY

Hence,
1 1
lim =
x c
x‚Ü’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‚Ü’c

Let > 0 be given. Since f is continuous at g(c), there exists őī1 > 0 such
that
1. |f (y) ‚ą’ f (g(c))| < , whenever, |y ‚ą’ g(c)| < őī1 .
Since g is continuous at c, and őī1 > 0, there exists őī > 0 such that
2. |g(x) ‚ą’ g(c)| < őī1 , whenever, |x ‚ą’ c| < őī.
On replacing y by g(x) in equation (1), we get

|f (g(x)) ‚ą’ f (g(c))| < , whenever, |x ‚ą’ c| < őī.

By deÔ¬Ānition, it follows that

lim f (g(x)) = f (g(c))
x‚Ü’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 ‚ą’ 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‚Ü’c

Let > 0 be given. Then > 0. Since f is continuous at c and > 0, there
2 2
exists some őī1 > 0 such that

|f (x) ‚ą’ f (c)| < , whenever, |x ‚ą’ c| ‚Č¤ őī1 .
(1)
2

Also, since g is continuous at c and > 0, there exists some őī2 > 0 such that
2
őī
|g(x) ‚ą’ g(c)| < , whenever, |x ‚ą’ c| < .
(2)
2 2
Let őī = min{őī1 , őī2 }. Then őī > 0. Let |x ‚ą’ c| < őī. Then |x ‚ą’ c| < őī1 and
|x ‚ą’ c| < őī2 . For this choice of x, we get

|{f (x) + g(x)} ‚ą’ {f (c) + g(c)}|
= |{f (x) ‚ą’ f (c)} + {g(x) ‚ą’ g(c)}|
‚Č¤ |f (x) ‚ą’ f (c)| + |g(x) ‚ą’ g(c)| (by triangle inequality)
< +
2 2
=.

It follows that
lim (f (x) + g(x)) = f (c) + g(c)
x‚Ü’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, őī1 , őī2 and őī exactly as in Part (i).
Suppose |x ‚ą’ c| < őī. Then |x ‚ą’ c| < őī1 and |x ‚ą’ c| < őī2 . For these choices of
x we get

|{f (x) ‚ą’ g(x)} ‚ą’ {f (c) ‚ą’ g(c)}|
= |{f (x) ‚ą’ f (c)} ‚ą’ {g(x) ‚ą’ g(c)}|
‚Č¤ |f (x) ‚ą’ f (c)| + |g(x) ‚ą’ g(c)| (by triangle inequality)
< +
2 2
=.

It follows that
lim (f (x) ‚ą’ g(x)) = f (c) ‚ą’ g(c)
x‚Ü’c

and, hence, f ‚ą’ 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 őī > 0 such that

whenever, |x ‚ą’ c| < őī.
|f (x) ‚ą’ f (c)| < ,
|k|

If |x ‚ą’ c| < őī, then

|kf (x) ‚ą’ kf (c)| = |k(f (x) ‚ą’ f (c))|
= |k| |(f (x) ‚ą’ f (c)|
< |k| ¬·
|k|
=.

It follows that
lim kf (x) = kf (c)
x‚Ü’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‚Ü’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 őī1 > 0 such that
2
|f (x) ‚ą’ f (c)| < whenever, |x ‚ą’ c| < őī1 .
1

Also, since g is continuous at c and > 0, there exists őī2 > 0 such that
1

|g(x) ‚ą’ g(c)| < whenever, |x ‚ą’ c| < őī2 .
1

Let őī = min{őī1 , őī2 } and |x ‚ą’ c| < őī. For these choices of x, we get

|f (x)g(x) ‚ą’ f (c)g(c)|
= |(f (x) ‚ą’ f (c) + f (c))(g(x) ‚ą’ g(c) + g(c)) ‚ą’ f (c)g(c)|
= |(f (x) ‚ą’ f (c))(g(x) ‚ą’ g(c)) + (f (x) ‚ą’ f (c))g(c) + f (c)(g(x) ‚ą’ g(c))|
‚Č¤ |f (x) ‚ą’ f (c)| |g(x) ‚ą’ g(c)| + |f (x) ‚ą’ f (c)| |g(c)| + |f (c)| |g(x) ‚ą’ 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‚Ü’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) ‚Č¤ g(x) ‚Č¤ h(x) for all x in an open
interval containing c and

lim f (x) = lim h(x) = L.
x‚Ü’c x‚Ü’c

Then, show that,
lim g(x) = L.
x‚Ü’c

Let > 0 be given. Then there exist őī1 > 0, őī2 > 0, and őī = min{őī1 , őī2 }
such that

whenever 0 < |x ‚ą’ c| < őī1
|f (x) ‚ą’ L| <
2
|h(x) ‚ą’ L) < whenever 0 < |x ‚ą’ c| < őī2 .
2
If 0 < |x ‚ą’ c| < őī1 , then 0 < |x ‚ą’ c| < őī1 , 0 < |x ‚ą’ c| < őī2 and, hence,

‚ą’ < f (x) ‚ą’ L < g(x) ‚ą’ L < h(x) ‚ą’ L < .
2 2
It follows that

|g(x) ‚ą’ L| < < whenever 0 < |x ‚ą’ c| < őī,
2
and
lim g(x) = L.
x‚Ü’c

Example 2.1.20 Show that f (x) = |x| is continuous at 0.
We need to show that
lim |x| = 0.
x‚Ü’0

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

lim |x| = 0
x‚Ü’0
2.1. INTUITIVE TREATMENT AND DEFINITIONS 57

Example 2.1.21 Show that
(i) lim sin őł = 0 (ii) lim cos őł = 1
őł‚Ü’0 őł‚Ü’0
1 ‚ą’ cos őł
sin őł
(iii) lim =1 (iv) lim =0
őł őł
őł‚Ü’0 őł‚Ü’0

graph

Part (i) By deÔ¬Ānition, the point C(cos őł, sin őł), where őł is the length of
the arc CD, lies on the unit circle. It is clear that the length BC = sin őł is
less than őł, the arclength of the arc CD, for small positive őł. Hence,
‚ą’őł ‚Č¤ sin őł ‚Č¤ őł
and
lim sin őł = 0.
őł‚Ü’0+
For small negative őł, we get
őł ‚Č¤ sin őł ‚Č¤ ‚ą’őł
and
lim sin őł = 0.
őł‚Ü’0‚ą’
Therefore,
lim sin őł = 0.
őł‚Ü’0

Part (ii) It is clear that the point B approaches D as őł tends to zero. There-
fore,
lim cos őł = 1.
őł‚Ü’0

Part (iii) Consider the inequality
Area of triangle ABC ‚Č¤ Area of sector ADC ‚Č¤ Area of triangle ADE
1 1 sin őł
1
cos őł sin őł ‚Č¤ őł ‚Č¤ .
2 2 2 cos őł
58 CHAPTER 2. LIMITS AND CONTINUITY

Assume that őł is small but positive. Multiply each part of the inequality by
2/ sin őł to get
őł 1
cos őł ‚Č¤ ‚Č¤ .
sin őł cos őł
On taking limits and using the squeeze theorem, we get
őł
lim = 1.
sin őł
+
őł‚Ü’0

By taking reciprocals, we get
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