Chapter 3: Continuous Functions
1. Limit of function:
Limit of function, left and right limit, operations, squeezing principle, monotone
convergence theorem, Cauchy convergence theorem, some important limits.
2. Continuity:
Continuity, operations and composition of continuous functions,
3. Extreme value theorem:
Maximizer and minimizer. Extreme value theorem. Compactness.
4. Intermediate value theorem.
5. Uniform continuity.
6. Inverse function.
7. Continuity of elementary functions:
Power function with rational exponential. Exponential function. Logarithm
function. Power function with real exponential.
1
1.1
Limit of Function
Limit
Limit point. A point p is called a limit point of the set S if every neighborhood of
p contains a point q ̸= p such that q ∈ S.
Definition 3.1.1 (Limit of a function) Let f be a function defined in the domain
D and x0 be a limit point of D. We say that the limit of function f is l as x tends to
x0 , provided that for any given ϵ > 0, there exists a δ > 0, such that for x ∈ D,
|f (x) − l| < ϵ, ∀ 0 < |x − x0 | < δ.
In this case, we denote
lim f (x) = l.
x→x0
Left limit and right limit If x is restricted from left (right) hand side tending to
x0 , i.e., x < x0 (x > x0 ), then the limit l is called left (right) limit, denoted by
(
)
lim f (x) = l
lim f (x) = l
x→x0 −0
x→x0 +0
It is easy to show that the limit exists if and only if both the left and right limits
exist and equal.
Theorem 3.1.2 Let f be a function defined in D and x0 is a limit point of D. Then
lim f (x) exists if and only if lim f (xn ) exists, where {xn } is an arbitrary sequence
x→x0
n→∞
in D and xn → x0 .
Proof. The necessary part: Assume lim f (x) = l, we want to show lim f (xn ) = l,
x→x0
n→∞
for any sequence xn → x0 . Since lim f (x) = l, then for given ϵ > 0, there exists a
x→x0
δ > 0, such that for x ∈ D,
|f (x) − l| < ϵ, if |x − x0 | < δ.
Since xn → x0 , then for δ > 0, there exists a N > 0, such that,
|xn − x0 | < δ
∀n ≥ N.
Thus we have
|f (xn ) − l| < ϵ,
∀n ≥ N,
i.e., lim f (xn ) = l.
n→∞
The sufficient part: Assume lim f (xn ) = l, for some sequence xn → x0 , and let
n→∞
yn → x0 be any sequence. Since both {xn } and {yn } can be subsequences of another sequence {zn } (e.g., take terms from {xn } and {yn } in turn.), lim f (zn ) exists,
n→∞
and f (zn ) → l, then f (yn ) → l. We want to show lim f (x) = l. We argue by conx→x0
tradiction. Suppose the conclusion is not true, which means that for some ϵ0 > 0, for
any δ > 0 we have
|f (x) − l| ≥ ϵ0 , for some x satisfying |x − x0 | < δ.
Hence for δn =
1
n
there is a point an such that
|an − x0 | < δn , but |f (an ) − l| ≥ ϵ0 .
2
Thus we obtain a sequence {an } in D that converges to x0 but {f (an )} does not
converge to l. This contradicts with the assumption.
Examples 3.1.1 Show that
lim x sin
x→0
1
= 0.
x
Proof. Let {xn } be any sequence with xn ̸= 0 and lim xn = 0. Since
n→∞
1 0 ≤ xn sin ≤ |xn | ,
xn
By Squeezing Principle
1 lim xn sin = 0.
n→∞ xn
Then from Theorem 3.1.2 we get lim x sin
x→0
1
= 0,
x
Examples 3.1.2 Show that
)x
(
1
lim 1 +
= e.
x→∞
x
)n
(
1
lim 1 +
= e,
n→∞
n
Proof. We have defined
where n is a natural number. Let {kn } be any subsequence of {n}, then
(
)k
1 n
lim 1 +
= e.
n→∞
kn
Now for an arbitrary sequence {xn }, xn → ∞, we prove
(
)xn
1
lim 1 +
= e.
n→∞
xn
Assume all xn > 1, let kn = [xn ], we have
1+
1
1
1
<1+
<1+ .
kn + 1
xn
kn
Then we have
(
1+
1
kn + 1
)kn
(
)xn (
)k +1
1 n
1
< 1+
.
< 1+
xn
kn
3
Since
(
lim 1 +
n→∞
1
kn + 1
(
)k +1
1 n
lim 1 +
n→∞
kn
By Squeezing Principle
)kn +1
1 + kn1+1
e
) = = e,
= lim (
n→∞
1
1 + kn1+1
)k (
)
(
1 n
1
· 1+
= e · 1 = e.
= lim 1 +
n→∞
kn
kn
(
)kn
(
)xn
1
lim 1 +
= e.
n→∞
xn
Since {xn } is an arbitrary sequence such that xn → ∞, therefore
(
)x
1
lim 1 +
= e.
x→∞
x
Examples 3.1.3 Let f (x) =
√
x, x ≥ 0. Show that
√
lim
x = 0.
x→0+0
Proof. {xn } be any sequence with xn ̸= 0 and lim xn = 0. Hence for any give ϵ > 0,
n→∞
let ϵ1 = ϵ2 and choose N such that xn < ϵ1 , ∀n ≥ N , then we have
√
xn <
√
ϵ1 = ϵ, ∀n ≥ N.
The result follows from Theorem 3.1.2.
Examples 3.1.4 Prove that
lim
x→0
sin x
= 1.
x
Proof. From figure 3.1.1, the areas of the triangle OCB, the sector OCB, and the
triangle OCD satisfy
1
1
1
|OC||AB| ≤ |OC|2 |x| ≤ |OC||CD|,
2
2
2
4
|x| <
π
.
2
D
B
x
O
A
C
Figure 3.1.1
Let |OC| = 1, then |AB| = | sin x|, |CD| = | tan x|. Hence,
| sin x| ≤ |x| ≤ | tan x|,
Then we have
| cos x| =
Since cos2 x =
√
|x| <
| sin x|
| sin x|
≤
≤ 1,
| tan x|
|x|
π
.
2
|x| <
π
.
2
1 − sin2 x and | sin x| < |x|, then
√
| cos x| ≥ 1 − |x|2 , |x| ≤ 1.
Since sin x and x have a same sign wher |x| ≤ 1,
√
1 − |x|2 ≤
sin x
≤ 1,
x
|x| ≤ 1.
Let {xn } be any sequence satisfying xn ̸= 0 and lim xn = 0. Then there is a natural
n→∞
number N such that
|xn | ≤ 1, ∀n ≥ N.
Hence
√
sin xn
1 − |xn |2 ≤
≤ 1,
xn
By Squeezing Principle,
sin xn
= 1.
n→∞ xn
lim
Therefore,
sin x
= 1.
x→0 x
lim
5
n ≥ N.
Definition 3.1.3 (Limit at infinity) Let f(x) be defined on (a, +∞). We say that
f (x) converges to l as x tends to +∞ if for given ϵ > 0, there exists a constant M
such that
|f (x) − l| < ϵ, ∀x ≥ M,
and denote by
lim f (x) = l.
x→+∞
Similar to the limit of sequences, we can define other types of limit.
1.2
Operations and properties
Similar to the limit of sequences, we have the following properties for the limit of
functions.
Theorem 3.1.4 Suppose that
lim f (x) = a, and lim g(x) = b.
x→x0
x→x0
Then
(i) lim [(f (x) ± g(x)] = lim f (x) ± lim g(x) = a ± b.
x→x0
x→x0
x→x0
(ii) lim [f (x) · g(x)] = lim f (x) · lim g(x) = a · b.
x→x0
x→x0
x→x0
lim f (x)
f (x)
a
x→x0
(iii) If g(x) ̸= 0, ∀x, and b ̸= 0, then, lim
=
= .
x→x0 g(x)
lim g(x)
b
x→x0
Theorem 3.1.5 (Squeezing Principle) Suppose that f (x) ≤ g(x) ≤ h(x), and
lim f (x) = lim h(x) = a.
x→x0
x→x0
then lim g(x) = a.
x→x0
Definition 3.1.6 The function f : D → R is called monotonically increasing provided that
f (v) ≥ f (u), ∀u, v ∈ D s.t. v > u.
The function f : D → R is called monotonically decreasing provided that
f (v) ≤ f (u),
∀u, v ∈ D s.t. v > u.
6
In both cases, if the equality does not hold, then the function is called strictly monotone
(strictly increasing or strictly decreasing).
Theorem 3.1.7 (Monotone Convergence Theorem) Suppose that f (x) is defined
on some interval (a, b) and monotonically increasing. If f (x) is bounded above, i.e.,
f (x) ≤ M
∀x ∈ (a, b),
then lim f (x) exists.
x→b−
Theorem 3.1.8 (Cauchy Convergence Criterion) lim f (x) exists if and only if
x→x0
for given ϵ > 0, ∃δ > 0, such that for any x and x′ satisfying 0 < |x − x0 | < δ and
0 < |x′ − x0 | < δ, we have
|f (x) − f (x′ )| < ϵ.
2
2.1
Continuity
Definition
Let f be a function defined in the domain D and x0 ∈ D is a limit point of D.
Definition 3.2.1 (Continuity) The function f : D → R is said to be continuous
at the point x0 ∈ D provided that
lim f (x) = f (x0 ).
x→x0
In terms of ϵ − δ: For any given ϵ > 0, there exists a δ > 0, such that for x ∈ D,
|f (x) − f (x0 )| < ϵ,
∀|x − x0 | < δ.
The function f : D → R is said to be continuous in D provided that it is continuous
at every point in D.
2.2
Operations and composition on continuous functions
Theorem 3.2.2 Suppose that the functions f :
continuous in D. Then
(i) f (x) ± g(x) is continuous in D;
(ii) f (x) · g(x) is continuous in D;
7
D → R and g :
D → R are
(iii) if g(x) ̸= 0 in D, then
f (x)
is continuous in D.
g(x)
Theorem 3.2.3 (Continuity of composition) Suppose that f : D → R is continuous at the point x0 ∈ D and g : U → R is continuous at the point f (x0 ) ∈ f (D) ⊆ U .
Then the composition g ◦ f is continuous at x0 .
Proof. Let {xn } be a sequence in D and xn → x0 . By the continuity of f at point
x0 , we get f (xn ) → f (x0 ). Then {f (xn )} is a sequence in U and converges to f (x0 ),
so by the continuity of g at point f (x0 ) we get
g(f (xn )) → g(f (x0 )),
which means that g ◦ f is continuous at x0 .
2.3
Examples
Examples 3.2.1 The function
Pn (x) =
n
∑
ak xn−k
k=0
is continuous everywhere.
Proof. It is obvious that f (x) = x is continuous in (−∞, ∞). Hence, by the theorem
of operations on continuous functions, Pn (x) is continuous in (−∞, ∞).
Examples 3.2.2 f (x) = sin x is continuous everywhere.
Proof. For any x0 ∈ (−∞, ∞)
sin x − sin x0 = 2 sin
x − x0
x − x0
cos
.
2
2
It has been proved that | sin x| ≤ |x|, then
x − x0 ≤ 2 |x − x0 | = |x − x0 |.
| sin x − sin x0 | ≤ 2 sin
2 2
Hence, ∀ϵ > 0, choose δ = ϵ, then
| sin x − sin x0 | ≤ |x − x0 | < ϵ, if |x − x0 | < δ,
which means that f (x) is continuous at x0 .
8
Examples 3.2.3 f (x) =
√
x is continuous in [0, ∞).
Proof. It has been proved f (x) is continuous at the point 0. For x0 > 0, we have
√
√
|x − x0 |
|x − x0 |
| x − x0 | = √
≤ √
.
√
x0
| x + x0 |
∀ϵ > 0, choose δ =
√
x0 ϵ, then
√
√
| x − x0 | < ϵ, if |x − x0 | < δ,
which means that f (x) is continuous at x0 .
(
)
Examples 3.2.4 cos x is continuous everywhere; tan x is continuous in − π2 , π2 .
√
Proof. Proof: Let f (x) = 1 − sin2 x and g(x) = x. f (x) is continuous in (−∞, ∞),
g(x) is continuous in [0, ∞), and f ((−∞, ∞)) ⊆ D(g). Since
{ √
[ π
]
2
π
1
−
sin
x
=
g
◦
f
(x),
x
∈
−
+
2kπ,
+
2kπ
2
√
[ 2
]
cos x =
− 1 − sin2 x = −g ◦ f (x),
x ∈ π2 + 2kπ, 3π
+ 2kπ ,
2
k = 0, 1, · · ·. By the theorem of composition on continuous functions, cos x is continuous everywhere. Since
( π π)
sin x
tan x =
, x∈ − ,
,
cos x
2 2
(
)
it is continuous in − π2 , π2 due to the continuity of sin x and cos x.
Examples 3.2.5 The function
{
f (x) =
sin x
,
x
1,
x ̸= 0,
x=0
is continuous everywhere.
Proof. By the theorem of operations on continuous functions,
everywhere except x = 0. We have proved that
lim
x→0
sin x
= 1.
x
Since f (0) = 1, we have
lim f (x) = f (0),
x→0
i.e., f (x) is continuous at x = 0.
9
sin x
x
is continuous
Examples 3.2.6 Let
x2n − 1
,
n→∞ x2n + 1
then f (x) is not continuous (discontinuous) at x = ±1.
f (x) = lim
Examples 3.2.7 Let
{
f (x) =
1
,
x3
0,
x ̸= 0,
x = 0,
then f (x) is not continuous at x = 0.
Examples 3.2.8 Let X : R → R (Dirichlet function) be defined by
{
1,
x is rational,
X (x) =
0,
x is irrational,
then X (x) is not continuous everywhere.
Proof. In case x0 is a rational number, X (x0 ) = 1. Let ϵ0 = 21 . By the density
theorem, no matter what the δ is chosen, the interval (x0 − δ, x0 + δ) always contains
an irrational number x̄. Since X (x̄) = 0, we have
|X (x̄) − X (x0 ) = 1 > ϵ0 ,
i.e., X (x) is not continuous at x0 . Similarly, we can prove the case of an irrational
number.
3
Extreme Value Theorem
In sections 2-4, we will apply the theory mentioned in the first two chapters to prove
three important theorems about continuous functions defined in a bounded closed
interval.
3.1
Maximizer and minimizer
Definition 3.3.1 We say that a function f : D → R attains a maximum (minimum)
provided that its image f (D) has a maximum (minimum); that is, there is a point x0
(not necessary only one) in D such that
f (x) ≤ f (x0 ) (f (x) ≥ f (x0 )),
∀x ∈ D.
We call such a point in D a maximizer (minimizer) of the function f .
In general, no assertion can be made concerning the existence of a maximum or
minimum. But we have the following important theorem about a continuous function
defined on a bounded closed interval [a, b].
10
3.2
Extreme value theorem
Theorem 3.3.2 (Extreme Value Theorem) Suppose that f : [a, b] → R is continuous. Then f attains both a minimum and a maximum in [a, b].
We first to prove a weaker result as the following lemma.
Lemma 3.3.3 Suppose that f : [a, b] → R is continuous. Then the image of f :
[a, b] → R is bounded; that is, there is a number M such that
|f (x)| ≤ M,
∀x ∈ [a, b].
Proof. We will argue by contradiction. If f ([a, b]) is not bounded above, then for
every natural number n, there must be a point xn such that
n = 1, 2, 3, · · ·
f (xn ) > n,
Since a ≤ xn ≤ b, by Bolzano-Weierstrass Theorem, there is a subsequence {xnk } of
{xn } converges to a point x0 ∈ [a, b]. Hence on one hand, f (xnk ) > nk i.e. {f (xn )}
has no upper bound; on the other hand, due to continuity of f, f (xnk ) → f (x0 ).
This contradiction show that f ([a, b]) is bounded above. Similarly we can prove
that f ([a, b]) is bounded below. If it is not true, then we may choose a sequence {xn }
such that
f (xn ) < −n, n = 1, 2, 3, · · ·
Then we have a convergent subsequence {xnk } that on one hand {f (xn )} has no lower
bound and on the other hand f (xnk ) → f (x0 ) as xnk → x0 .
Proof of Theorem 3.3.2 Since f ([a, b]) is bounded above, according to the Completeness Axiom, f ([a, b]) has the least upper bound. Define C = sup f ([a, b]). Now,
we want to find a point x0 ∈ [a, b] such that f (x0 ) = C. Then
f (x0 ) = max f [a, b],
or,
f (x0 ) ≥ f (x),
∀x ∈ [a, b].
Since C is the least upper bound of f ([a, b]), for any natural number n, C −
an upper bound. Hence there is a sequence of points xn at which
f (xn ) > C −
Then we have
C−
1
,
n
n = 1, 2, 3, · · ·
1
< f (xn ) ≤ C,
n
11
1
n
is not
it follows that the sequence {f (xn )} converges to C. Again according to BolzanoWeierstrass theorem, there is a subsequence {xnk } of {xn } that converges to a point
x0 ∈ [a, b]. Since f is continuous on [a, b], {f (xnk )} converges to f (x0 ). But {f (xnk )}
is a subsequence of {f (xn )}, so the limit of {f (xn )} is also the limit of {f (xnk )},
i.e. f (x0 ) = C. Thus we have proved that x0 is a maximizer and f (x0 ) attains the
maximum.
Because f ([a, b]) is bounded below, f ([a, b]) has a greatest low bound. In a similarly way by Bolzano-Weierstrass theorem we can find a minimizer at which f attains
a minimum.
Note: If a continuous function defined in an open interval. It may not have maximum
and minimum. The reason is that if lim xn = x0 and a < xn < b, the limit x0 may
n→∞
be the end point a or b.
Example 3.3.1 Let f (x) = x, 0 < x < 1. The function f is continuous in (0, 1), but
it has no maximum and minimum in the interval.
3.3
Compactness
Definition 3.3.4 A set S is said to be compact (sequentially compact) provided that
every sequence in S has a subsequence that converges to a point of S.
In this new terminology, the Bolzano-Weierstrass Theorem is simply the assertion
that the set of a bounded closed interval [a, b] is compact.
Theorem 3.3.5 Suppose that S is a compact nonempty set of real numbers and
f : K → R is continuous. Then f : S → R attains both a minimum and a
maximum.
4
Intermediate Value Theorem
Theorem 3.4.1 Suppose that the function f : [a, b] → R is continuous, and
f (a) < 0,
f (b) > 0.
Then there is at least a point x0 in the open interval (a, b) at which f (x0 ) = 0.
Proof. Let a1 = a, b1 = b and c1 = 12 (a1 + b1 ).
If f (c1 ) ≤ 0, define a2 = c1 and b2 = b1 .
If f (c1 ) > 0, define a2 = a1 and b2 = c1 .
In general, let cn = 12 (an + bn ).
12
If f (cn ) ≤ 0, define an+1 = cn and bn+1 = bn .
If f (cn ) > 0, define an+1 = an and bn+1 = cn .
Hence we define a sequence of nested, closed intervals {[an , bn ]} that satisfies f (an ) ≤
0, f (bn ) > 0,
a ≤ an ≤ an+1 < bn+1 ≤ bn ≤ b,
and
1
1
bn − an = (bn−1 − an−1 ) = · · · = n−1 (b − a).
2
2
Thus the sequences {an } and {bn } satisfy the assumptions of the Nested Interval
Theorem, so there is a point x0 ∈ [a, b] such that an → x0 and bn → x0 . Since f (x)
is continuous on [a, b],
lim f (an ) = lim f (bn ) = f (x0 ).
n→∞
n→∞
Since f (an ) ≤ 0, ∀n, we have f (x0 ) ≤ 0, and on the other hand we have f (x0 ) ≥ 0
due to f (bn ) ≥ 0, ∀n. Hence f (x0 ) = 0.
Theorem 3.4.2 (Intermediate Value Theorem) Suppose that the function f :
[a, b] → R is continuous and c is a number strictly between f (a) and f (b), i.e.,
f (a) < c < f (b), or f (a) > c > f (b).
Then there is a point x0 in the open interval (a, b) at which f (x0 ) = c.
Proof. First let us suppose that f (a) < c < f (b). Define the function g(x) = f (x)−c.
Then
g(a) < 0 < g(b).
It follows from Theorem 3.3.1 that there is a point x0 ∈ (a, b) at which g(x0 ) = 0, i.e.
g(x0 ) = f (x0 ) − c = 0, or f (x0 ) = c.
In the case when f (a) > c > f (b), define g(x) = c − f (x) and follow the same
argument.
5
Uniform Continuity
Suppose that a function f : D → R is continuous in D. According to the ϵ − δ
definition of continuity, for x0 ∈ D and given ϵ > 0, there exists a δ > 0, such that
for x ∈ D,
|f (x) − f (x0 )| < ϵ, ∀|x − x0 | < δ.
13
In general, δ depends not only on ϵ, but also on x0 , i.e., at different points, we need
to choose different δ such that the above inequality is true. Sometimes the δ may
need to tend to 0 even for a fixed ϵ, for example, f (x) = 1/x, x ∈ (0, 1). Then we
may ask the question: What extra conditions are needed in order to find a δ which
is good for any x0 ∈ D?
Definition 3.5.1 (Uniformly continuity) Suppose that the function f (x) is defined
on an interval D. f (x) is said to be uniformly continuous in D provided that for any
given ϵ > 0, ∃δ > 0, s.t.
|f (u) − f (v)| < ϵ,
∀u, v ∈ D, |u − v| < δ.
If D is an open interval, f : D → R is continuous in D, but f may not be
uniformly continuous in D.
Example 3.5.1 Let f (x) = x1 , D = (0, 1), then f (x) is continuous in D, but not
uniformly continuous in D. Since for any u, v ∈ D,
1 1 u − v − =
u v uv .
Take ϵ = 1, then for any δ > 0, let 0 < u < δ and v = u2 , we have |u − v| =
but
1 1 1
|f (u) − f (v)| = − = > 1.
u v
u
If D is a bounded closed interval, we have the following theorem.
u
2
< δ,
Theorem 3.5.2 A continuous function on a closed bounded interval, f : [a, b] → R,
is uniformly continuous.
Proof. We will argue by contradiction. Let D = [a, b]. Suppose that f : D → R is
not uniformly continuous. Then there is some ϵ0 > 0 such that ∀δ > 0
|f (u) − f (v)| ≥ ϵ0 ,
for some u, v ∈ D, |u − v| < δ.
Let δ = n1 , then there exist points un and vn in [a, b] such that |un − vn | < n1 , but
|f (un ) − f (vn )| ≥ ϵ0 ,
∀n.
This defines two sequences {un } and {vn } in [a, b]. By using Bolzano-Weierstrass
Theorem, we can choose a subsequence {unk } of {un } that converges to a point
u∗ ∈ [a, b]. But for each natural number k,
|unk − vnk | <
14
1
1
≤ ,
nk
k
or
1
1
< vnk < unk + .
k
k
∗
By Squeeze Principle, {vnk } also converges to u . The assumption of continuity of f
at point u∗ implies that both {f (unk )} and {f (vnk )} converge to f (u∗ ). Thus their
difference {f (unk ) − f (vnk )} converges to 0. This contradicts with the inequality
|f (unk ) − f (vnk )| ≥ ϵ0 . It follows that f : D → R is uniformly continuous.
unk −
6
Inverse Function
Definition 3.6.1 A function f : D → R is said to be one-to-one provided that for
each point y in its image f (D), there is exactly one point x in its domain D such that
f (x) = y. In other words, if f (x1 ) = f (x2 ), then x1 = x2 . For a one-to-one function
f : D → R, we define its inverse, denoted by
f −1 : f (D) → R,
as following:
f −1 (y) = x, if f (x) = y.
It is clear that if f : D → R is strictly monotone, it is one-to-one and its inverse
f : f (D) → R is also strictly monotone. In fact, if f is strictly increasing, then
f −1 is also strictly increasing; if f is strictly decreasing, then f −1 is also strictly
decreasing.
−1
Theorem 3.6.2 Let D be an interval and suppose that the function f : D → R is
monotone. If its image f (D) is an interval, then the function f is continuous.
Proof. Since −f is increasing if f is decreasing, we only need to consider the case
when f is increasing. For any x0 ∈ D, it is easy to show that lim f (x) = f (x0 − 0)
x→x0 −0
and
lim f (x) = f (x0 + 0) exist. Suppose that f (x) is not continuous at x0 , for
x→x0 +0
example, f (x0 − 0) < f (x0 ). Since f (x) ≤ f (x0 − 0) when x < x0 and f (x) ≥ f (x0 )
when x > x0 , then f (x) can not take the values between f (x0 − 0) and f (x0 ), which
contradicts with the assumption that f (D) is an interval.
Theorem 3.6.3 Let I be an interval and suppose that the function f : I → R is
strictly monotone. If f (I) is an interval, then the inverse function f −1 : f (I) → R
is continuous.
Proof. Suppose that f is strictly increasing. Let D = f (I). Then the function
f −1 : D → R is a monotonically increasing function that has the interval I as its
image. From Theorem 3.6.3, f −1 : D → R is continuous.
15
Theorem 3.6.4 Let I be an interval and the function f : I → R is strictly monotone
and continuous. Then the inverse function f −1 : f (I) → R is continuous and the
definition domain f (I) is an interval.
Proof. Let y1 = f (x1 ), y2 = f (x2 ) ∈ f (I), x1 < x2 , and c be any number in
between y1 and y2 . Then from the Intermediate Value Theorem there exits a point
x0 ∈ (x1 , x2 ) such that f (x0 ) = c. Hence c ∈ f (I), i.e., f (I) is an interval. Since f (I)
is an interval, by Theorem 3.6.3. f −1 is continuous.
Example 3.6.2 Define f : [0, ∞) → R, as y = f (x) = x2 . The inverse x = f −1 (y) =
√
√
y. We have proved that the function y is continuous in [0, ∞) by the definition
of continuity. Now, we apply Theorem 3.6.4 to prove this. Because f (x) is strictly
increasing and continuous in [0, ∞) and the image of f is f ([0, ∞)) = [0, ∞). Hence
√
by the theorem, f −1 (y) = y is continuous in [0, ∞).
Example 3.6.2 Inverse functions sin−1 x, cos−1 x and tan−1 x.
We have proved
sin x is strictly
[ π π ]the continuity of sin x, cos x and tan x. Since
−1
increasing in − 2 , 2 and the image is [−1, 1], hence its inverse sin x is continuous
in [−1, 1] and also strictly increasing.
Because cos x is strictly decreasing in [0, π] and the image is [−1, 1], hence its
inverse cos−1 x is continuous in [−1, 1] and[ also strictly
decreasing.
]
Because tan x is strictly increasing in − π2 , π2 and the image is (−∞, ∞). Hence
its inverse tan−1 x is continuous on (−∞, ∞) and also strictly increasing.
7
Continuity of Elementary Functions
In section 1 and 6, we have seen the continuity of polynomials Pn (x), the trigonometric functions sin x, cos x, and tan x and their inverses. Using the properties of
the continuous functions, we can easily show that the rational functions and other
trigonometric functions are all continuous in their domains. In this section, we discuss
the continuity of some other elementary functions, such as power functions, logarithmic functions, and exponential functions. Then from the theorems for operations on
continuous functions, for inverse functions, and for compositions, we know that any
combinations and operations (including inverse and composition) on these functions
result in also continuous functions.
7.1
Power Function with Rational Exponential
1
• y = x n , x ≥ 0, n is a natural number.
Since x = y n is a strictly increasing function in interval [0, ∞) and its image is
1
also interval [0, ∞), its inverse function y = x n is continuous in [0, ∞).
16
m
• y = x n , x ≥ 0, m, n are natural numbers.
1
This function can be treated as a composite function of z = x n with y = z m .
m
Hence y = x n is continuous in [0, ∞).
7.2
Exponential Function
• y = ax , a > 0.
Function y = aγ has been defined when γ is rational. Let’s temporarily assume
a > 1. We consider f (γ) = aγ in the rational set Q. Then f (γ) has the following
properties.
(i) aγ1 +γ2 = aγ1 · aγ2
(ii) f (γ) is strictly increasing, i.e., aγ1 < aγ2 , if γ1 < γ2 .
(iii) For any sequence {γn } in Q that converges to 0, the sequence {f (γn )}
converges to a0 = 1.
(iv) If {γn } is a convergent sequence, {aγn } is a Cauchy sequence.
• Definition of ax . For any real number x, let {γn } be a rational sequence and
converges to x. It is known from the above property (iv) that {aγn } is a Cauchy
sequence. We define
ax = lim aγn .
n→∞
For this definition, we need to prove that for any other rational sequence {γn′ }
that converges to x,
′
lim aγn = lim aγn .
n→∞
Indeed
′
)
′
aγn − aγn + lim aγn
n→∞
n→∞
( ′
)
γn
γn −γn
= lim a
a
− 1 + lim aγn
n→∞
n→∞
( ′
)
= lim aγn · lim aγn −γn − 1 + lim aγn
lim aγn =
n→∞
n→∞
(
lim
n→∞
=
n→∞
n→∞
lim aγn .
n→∞
Hence, for a > 1 we have define ax on the whole real axis, i.e. the interval
(−∞, ∞).
• Properties of Function ax
(i) For any real numbers x1 and x2 , ax1 +x2 = ax1 · ax2 .
17
(ii) For any real number x, ax > 0.
(iii) y = ax (a > 1) is strictly increasing.
Lemma 3.6.1 y = ax is continuous at x = 0.
Proof. We need to show that for any real sequence {xn } that converges to 0, {axn }
converges to a0 = 1. By the density property of rational numbers, we can choose two
rational sequences {γn } and {γn′ } that satisfy
xn −
1
1
< γn < xn < γn′ < xn + .
n
n
when a > 1, ax is strictly increasing, hence
′
aγn < axn < aγn .
′
Since γn → 0 and γn′ → 0 then aγn → 1 and aγn → 1. By using Squeezing Principle
lim axn = 1.
xn →0
Up to now, we have assumed that a > 1. If 0 < a < 1, we define
ax = (a−1 )−x ,
and if a = 1, we define
ax = 1.
Obviously, y = ax is strictly decreasing, continuous at x = 0, and satisfies (i), (ii).
Theorem 3.6.2 The exponential function y = ax (a > 0) is continuous on (−∞, ∞)
and strictly increasing (decreasing) when a > 1 (a < 1). Its image is the interval
(0, ∞)
Proof. Proof For any point x0 ∈ (−∞, ∞)
ax − ax0 = ax0 (ax−x0 − 1).
Since lim ax = 1, we get
x→0
lim (ax − ax0 ) = lim ax0 (ay − 1) = 0,
x→x0
y→0
which means that ax is continuous at x = x0 .
18
7.3
Logarithmic Function
y = loga x and y = ln x: It has been known that the function x = ay is a strictly
monotone and continuous function from (−∞, ∞) to (0, ∞). Now we define its inverse
by y = loga x (y = ln x for a = e), which is from (0, ∞) to (−∞, ∞). Since ax (ex )
is strictly monotone, its inverse loga x (ln x) is continuous and also strictly monotone
on (0, ∞).
ln(1 + x)
= 1.
x
Proof. By the definition of Logarithm,
Example 3.7.1 Show that lim
x→0
1
1
ln(1 + x) = ln(1 + x) x .
x
1
Hence, let f (x) = (1 + x) x and g(y) = ln y. Since
)y
(
1
1
x
lim (1 + x) = lim 1 +
= e,
x→0
y→∞
y
(
then
)
lim g(f (x) = g lim f (x) = g(e) = ln e = 1.
x→0
7.4
x→0
Power Function
y = xα (α is a real number.) : This function can be defined as a composition of
y = f (x) = α ln x with g(y) = ey , i.e., (g ◦ f )(x) = g(f (x)) = eα ln x . Since ln x is
continuous in (0, ∞) and the image is (−∞, ∞), ex is continuous in (−∞, ∞). Hence
by the theorem of continuity on composite functions, y = xα = eα ln x is continuous in
(0, ∞).
19