116
Chapter 4
Limits and Continuity
4.
Limits and Continuity
that is continuous at every irrational number and discontinuous at every rational num
ber in its domain. Such a function is given in Example 4.2.2(g). An example of an in
creasing function having the same properties will also be given in Section 4 of this
chapter.
4.1 Limit of a Function
4.2 Continuous Functions
4.3 Uniform Continuity
4.4 Monotone Functions and Discontinuities
4.11
Limit of a Function
The basic idea underlying the concept of the limii of a function f at a point p is to study
the behavior of f at points close to, but not equal to, p. We illustrate this with the fol
lowing simple examples. Suppose that the velocity v(ft /sec) of a falling object is given
as a function v = v(t) of time t. If the object hits the ground in 1 = 2 seconds, then
v(2) = 0. Thus to find the velocity at the time of impact, we investigate the behavior of
v(?) as approaches 2, but is not equal to 2. Neglecting air resistance, the function (1)
is given as follows:
S321, OSI 0, there exists a 8 > 0 for which
\(x) – L] 0, there exists a 8 >0 such that
f(x) E N.(L) for all x € En (N.(p)\{p}).
This is illustrated graphically in Figure 4.1.
rp
L+E
LO
LE
(a) Let E be a nonempty subset of R and let f, g, and h be functions on E defined by
f(x) = c(CER), 8(x) = x, and h(x) = x, respectively. If p is a limit point of E, then
lim f(x) = c, lim 8(x) = p.
lim h(x) = p?
These limits are also expressed as lim c = c, lim x = p, and lim r’ = p?.
Even though we may feel that these limits are obvious, they still have to be proved.
We illustrate the method of proof by using the definition to prove that lim h(x) = p 0,
we choose d = min{1, €/(2p1 + 1)}. With this choice of 8, if x E E with 0 < lxpl
< dwe first of all have fx  pl < 1, and therefore also
1h(x) – p’ < (21p1 + 1)x – pl < (2p1 + 1)
(2p1 + 1)
Thus lim x = p?
(b) For x # 2, let f(x) be defined by
2  4
f(x)
€
= €.
DSP p8
XP
Figure 4.1
lim f(x) = 2
=
x  2
The domain of f is E = (00, 2) U (2,00), and 2 is clearly a limit point of E. We now
show that lim f(x) = 4. For x # 2,
1x2  4
\f(x) 41
4
x  2
Thus given e > 0, the choice 8 = e works in the definition.
(c) Consider the following variation of (b). Let g be defined on R by
4 = 4x + 2 – 41 = 5x – 21.
r4
Remarks
(a) In the definition of limit, the choice of 8 for a given e may depend not only on e
and the function, but also on the point p. This will be illustrated in Example 4.1.2(g).
(b) If p is not a limit point of E, then for 8 sufficiently small, there do not exist any
x € E so that 0 < lx  pl < 8. Thus if p is an isolated point of E, the concept of the
limit of a function at p has no meaning.
(c) In the definition of limit, it is not required that pe E, only that p is a limit point of
E. Even if p E E, and f has a limit at p, we may very well have that
lim f(x) f(p).
This will be the case in Example 4.1.2(c).
(d) Let ECR and p a limit point of E. To show that a given function f does not have
a limit at p, we must show that for every LER, there exists an e > 0, such that for
every 8 > 0, there exists an x E E with 0 < px  pl < 8, for which
\(x) – L 26.
We will illustrate this in Example 4.1.2(e).
** 2,
8(x) =
XP
X2
2,
x = 2.
For this example, 2 is a point in the domain of g, and it is still the case that ling 8(x) = 4.
However, the limit does not equal 8(2) = 2. The graph of g is given in Figure 4.2.
(d) Let E = (1,0) U (0,00). For x e E, let h(x) be defined by
x + 1  1
h(x) =
X
4.1
Limit of a Function
119
120
Chapter 4
Limits and Continuity
8(x). x+2
1 →
VA
(2.8(2))
We will show that for this function, lim f(x) fails to exist for every p ER. Fix PER.
Let LER and let
max{\L – 11, \L}.
Suppose e = L – 11. By Theorem 1.5.2, for any 8 > 0, there exists an x EQ such
that 0 < 1p – x1 < 8. For such an x,
\(x)  L1 = 11 L1 = €.
If e = L), then by Exercise 6, Section 1.5, for any > 0, there exists an irrational
number * with 0 < [x  pl < 8. Again, for such an x, \(x)  L1 = €. Thus with
e as defined, for any 8 > 0, there exists an x with 0 < fx  pl < 8 such that
\f(x) – 4 2 €. Since this works for every L ER, lim f(x) does not exist.
(f) Let f: R → R be defined by
so, XEQ,
+
2
+
0
1P
Figure 4.2 Graph of 8
f(x) = x x¢ Q.
We claim that lim h(x) = {. This result is obtained as follows: For x + 0,
Vx+11 x'+ 1  Vx+1+
Vx+ 1 + 1
X
x+p
Then lim f(x) = 0. Since \f(x)] = [x] for all x, given e > 0, any 8,0 < 8 5 €, will
work in the definition of the limit. A modification of the argument given in (e) shows
that for any p = 0, lim f(x) does not exist. An alternative proof will be provided in
Example 4.1.5(b).
(g) Our final example shows dramatically how the choice of 8 will generally depend
not only on e, but also on the point p. Let E = (0,00) and let f: É → R be defined by
=
=
Hvati vetiti
 14
lavet
f(x) =
We will prove that for p € (0,00),
x(Vx + 1 + 1) Vx + 1 + 1
From this last term we now conjecture that h(x) = { as *→ 0. By the above,
1
1  Vx + 1
Vx+
x + 1 + 1 12(Vx+
+1 + 1)!
1  Vx + 1)(1 + V (x + 1)]
2(Vx + 1 + 1)2
+ 1 + 1)2
1 lxl
2(Vx+ 1 + 1)2
For x E E we have (Vx+ 1 + 1)? > 1, and thus
x 씨
X
=
1
1
lim
1PX
р
=
If x > p/2, then
fx  pl 2
<
хр
Therefore, given e > 0, let 8 = min{p/2, p?e/2}. Then if 0 < lx  pl < 8.x > p/2,
and
ale)  1 <
Given e > 0, let 8 = e. Then for all x e E with 0 < lxl < 8,
box 
<
<
<
The 8 as defined depends on both p and e. This suggests that any 8 that works for
a given p and e must depend on both p and e. Suppose on the contrary that for a given
€ > 0, the choice of 8 is independent of p € (0,00). Then with e = 1, there exists a
8 >0 such that
and thus lim h(x) = {.
(e) Let f be defined on R as follows:
0
1
.
COROLLARY If f has a limit at p, then it is unique.
Theorem 4.1.3 is often applied to show that a limit does not exist. If one can find
a sequence {Pn} with Pn → P, such that {f(Pn)} does not converge, then lim f(x) does
not exist. Alternatively, if one can find two sequences {Pn} and {rn} both converging 10
p, but for which
lim f(p) # lim f(n),
then again lim f(x) does not exist. We illustrate this with the following two examples.
This contradiction proves that the choice of 8 must depend on both p and e.
R100
00
**
4.1.5
EXAMPLES
Sequential Criterion for Limits
Our first theorem allows us to reduce the question of the existence of the limit of a func
tion to one concerning the existence of limits of sequences. As we will see, this result
will be very useful in subsequent proofs, and also in showing that a given function does
not have a limit at a point p.
(a) Let E = (0,00) and f(x)
that
sin(17x), E E. We use the previous theorem to show
lim sin
x
4.1.3 THEOREM Let E be a subset of R, p a limit point of E, and f a realvalued function
defined on E. Then
lim f(x) = L if and only if lim f(px) = L
for every sequence {Pm} in E, with Pr + p for all n, and lim pa = p.
does not exist. Let Pn = 2n Dr. Then
f(p.) = sin(2n + 1)= (1)".
Thus lim f(Pm) does not exist, and consequently by Theorem 4.1.3, lim f(x) also does
not exist. The graph of f(x) = sin(17x) is given in Figure 4.3.
+
noc
n00
XP
* *
Remark. Since p is a limit point of E, Theorem 2.4.7 guarantees the existence of a
sequence {Pm} in E with Pn p for all n E N and Pr → p.
Proof. Suppose lim f(x) = L. Let {Pn} be any sequence in E with Pr + p for all n
and Pr + p. Let e > 0 be given. Since lim f(x) = L, there exists a d > 0 such that
\f(x) – 4 0
4.1
Limit of a Function
123
124
Chapter 4
Limits and Continuity
1
100
(b) As in Example 4.1.2(f) let
so, xeQ,
f(x) =
lx, XEQ.
Suppose p ER, p = 0. Since Q is dense in R, there exists a sequence {Pn} cQ with
Po t p for all n EN such that po → p. Hence, lim f(Pn) = 0. On the other hand,
since RIQ is also dense in R, there exists a sequence {9n} of irrational numbers with
In → p. But then lim f(an) = lim qn = p. Thus since p = 0, by Theorem 4.1.3,
lim f(x) does not exist.
+00
the above &, there exists an n, E N such that 0 < \pm  pl < 8, for all n 2 no. Thus
8(Pn) 0 for all n 2 ng. Therefore by Theorem 2.2.1(c),
1
lim
8(P) B
Since this holds for every sequence Pr + p, by Theorem 4.1.3,
1 1
lim
**P 8(x) B
The proofs of the following two theorems are easy consequences of Theorem 4.1.3
and the corresponding theorems for sequences (2.2.3 and 2.2.4). First, however, we give
the following definition.
Limit Theorems
4.1.7
4.1.6 THEOREM Suppose E C R.3.8: E → R. and p is a limit point of E. If
lim f(x) = A and
lim 8(x) = B,
DEFINITION A realvalued function f defined on a set E is bounded an E if there
exists a constant M such that \f(x) = M for all x E E.
+P
x+p
then
4.1.8 THEOREM Suppose E C R, pis a limit point of E, and f, 8 are realvalued functions
on E. If 8 is bounded on E and lim f(x) = 0, then
lim f(x)g(x) =
= 0.
P
P
Proof. Exercise 12.
E
(a) lim (f(x) + 8(x)] = A + B,
(b) lim f(x)8(x) = AB, and
f(x) A
(c) lim
*++P8(x) B'
provided B +0.
Proof. For (a) and (b) apply Theorem 4.1.3 and Theorem 2.2.1. We leave the details
to the exercises (Exercise 11).
Proof of (c): By (b) it suffices to show that
4.1.9
THEOREM Suppose E C R, p is a limit point of E, and f, g, h are functions from E
into R satisfying
8(x) = f(x) sh(x) for all x E E.
If lim g(x) = lim h(x) = L, then lim f(x)
= L.
Proof. Exercise 13.
We now provide several examples to illustrate the previous theorems.
1 1
lim
** 8(x) B
We first show that since B # 0, 8(x) + 0 for all x sufficiently close to p, x # p. Take
€ = \B/2. Then by the definition of limit, there exists a 8, > such that
IBI
18(x) – Bl<
2
4.1.10
xc
for all x € 1,0 < 1*  pl < 81. By Corollary 2.1.4, 18(x) – B = 118(x)} – BIL.
Thus
EXAMPLES
(a) By Example 4.1.2(a), lim x = c. Thus, using mathematical induction and Theorem
4.1.6(b), lim x* = c" for all n E N. If p(x) is a polynomial function of degreen, that is,
p(x) = ant" + ... + apx + ans
where n is a nonnegative integer and an .. ., an E R with an # 0, then a repeated ap
plication of Theorem 4.1.6(a) gives lim p(x) = p(c).
(b) Consider
x + 2x  2x  4
lim
4
18(x) > BI
BI
>0
*+
2
for all x E E, O < 1x – pl < 8.
We can now apply Theorem 4.1.3 and the corresponding result for sequences of
Theorem 2.2.1. Let {Pn} be any sequence in E with Pn → p and Pr + p for all n. For
جر
4.1
Limit of a Function
127
128
Chapter 4
Limits and Continuity
(b) For our second example consider f(x) = x sin Tx. If we set Pn = = (n + 4), n EN
then
x+00
f(en) = (n + {) sin(n + 1)= (1)"(n + 2).
Thus the sequence {f(pm)} = 1 is unbounded, and as a consequence, lim x sin Tx does
not exist.
OC
x
for all x E Dom fn (M, 00). (See Figure 4.6.) If this is the case, we write
lim f(x) = L.
Similarly, if Dom fn(0, b) + for every b ER,
lim f(x) = 2
if and only if given e > 0, there exists a real number M such that
\(x)  L1 <
for all x € Domf n(0, M).
The hypothesis that Dom in (a,00) + 0 for every a E R is equivalent to saying
that the domain of the function f is not bounded above. If Dom f = N, then the above
definition gives the definition for the limit of a sequence. Readers should convince
themselves that all theorems up to this point involving limits at a point p E R are still
valid if p is replaced by oo or oo.
EXERCISES 4.1
X
=
1 + x
x + 1
1. Use the definition to establish each of the following limits.
*a. lim (2x  7) = 3
b.
lim, (3x + 5) = 1
1
*c. lin
d. lim 2r?  3x  4 = 1
2
 2x  4 5
e. lim
= 3
f. lim
1 x + 1
x  4 2
2. Use the definition to establish each of the following limits.
a. lim c = c
b. lim x = p
c. lim r' = p)
d. lim x" = p", "EN
Vx+pVe
e. lim Ve= Vo, p>0
1
f. lim
p>
1p
3. For each of the following, determine whether the indicated limit exists in R. Justify your answer!
r  1
*a. lim
b. lim
+ 1x1
3+1 x + 1
IP
IP
1P
0
X
Let
L
LE
X
0
M
x0
X
Figure 4.6 lim f(x) = L
+00
4.1.12
EXAMPLES
(a) As our first example, consider the function f(x) = (sin x)/x defined on (0,00).
Since (sin x) = 1,
*c. lim cos d.
cos
0 X
(x + 1)2 – 1
e. lim
1.
xx
1 \  11
4. *Define f:(1,1) R by
f(x) =
x x2
x +1
Determine the limit L of f at  1 and prove, using € and 8, that f. has limit L at  1.
5. *a. Using Figure 4.5, prove that sin hl sh for all h E R.
b. Using the trigonometric identity 1 – cos h = 2 sin? 2. prove that
(1) fim cos h = 1.
1  cos h
(ü) lim
= 0.
h
6. Let E CR, p a limit point of E, and f:E → R. Suppose there exist a constant M >0 and LE R such that
1()  L s Mbx  pl for all x E E. Prove that lim f(x) = L.
<
x
カー46
\
for all x € (0,00). Let e > 0 be given. Then with M = 1/€,
\f(x) M.
Therefore, lim (sin x)/x = 0.

100
4.1
Limit of a Function
129
130
Chapter 4
Limits and Continuity
x+p
19. Let f: R → R satisfy f(x + 1) = f(x) + f() for all x, y E R. If lim f(x) exists, prove that
a. lim f(x) = 0, and
limf(x) exists for every p ER.
b.
xp
xp
IP
4.2 Continuous Functions
7. Suppose f: E → Rp is a limit point of E, and lim f(x) = L.
*a. Prove that lim f(x)} = L.
b. If, in addition, f(x) > 0 for all x € E, prove that lim Vf(x) = VI.
*c. Prove that lim ((x))" = L" for each n E N.
8. Use the limit theorems, examples, and previous exercises to find each of the following limits. State which theo
rems, examples, or exercises are used in each case.
5x + 3x – 2
r  +2
a.
b.
X1
x+1
3x + 1
1x + 2132
d. lim
2x + 5
*+2 (x + 2)
1
*e. lim
1. lim
x  4
x+p
sin 2x
Ir  21  1x + 21
h. lim
*c. lim
Vx2
*g lim
X
x0
X
The notion of continuity dates back to Leonhard Euler (17071783). To Euler, a con
tinuous curve (function) was one that could be expressed by a single formula or equa
tion of the variable x. If the definition of the curve was made up of several parts, it was
called discontinuous. This definition was sufficient to convey the concept of continuity
in Euler's time as mathematicians were primarily concerned with elementary functions;
namely, functions built up from the trigonometric and exponential functions, and in
verses of these functions, using algebraic operations and composition.
The more modern version of continuity is credited to Bolzano (1817) and Cauchy
(1821). Both men were motivated to provide a clear and precise definition of continu
ity in order to prove the intermediate value theorem (Theorem 4.2.11). Cauchy's defin
ition of continuity was as follows: "The function f(x) will be, between two assigned val
ues of the variable x, a continuous function of this variable if for each value of x
between these limits, the numerical value (i.e., absolute value) of the difference
f(x + a) – f(x) decreases indefinitely with a“. Even this definition appears strange in
comparison with the more modern definition in use today. Both Bolzano and Cauchy
were concerned with continuity on an interval, rather than continuity at a point.
XP
4.2.1
9. *Suppose f:(a, b) → R, p € (a, b), and lim f() > 0. Prove that there exists a 8 >0 such that f(x) > 0 for all
xe (a, b) with 0 < br  pl < 8.
10. Suppose E C R, p is a limit point of E, and f: E → R. Prove that if f has a limit at p, then there exists a posi
tive constant M and a 8 > 0, such that \f(x) S M for all x E E, 0 < lx  pl < 8.
11. a. Prove Theorem 4.1.6(a).
b. Prove Theorem 4.1.6(b).
12. *Prove Theorem.4.1.8.
13. Prove Theorem 4.1.9.
14. La f, 8 be realvalued functions defined on ECR and let p be a limit point of E.
a. If lim f(x) and lim (f(x) + g(x)) exist, prove ihat lim 8(x) exists.
b. If limf(x) and lim (F(x)8(x)) exist, does it follow that lim 8(x) exists?
15. Let E be a nonempty subset of R and let p be a limit point of E. Suppose f is a bounded realvalued function on
E having the property that lim f(x) does not exist. Prove that there exist sequences {Pn} and {an} in E with
lim P* = lim qn = p such that lim (px) and lim f(an) exist, but are not equal.
16. *Let f be a realvalued function defined on (a,00) for some a > 0. Define 8 on (0, ) by 8(t) = s(). Prove that
lim f(x) = L if and only if lim 8(t) = L.
17. Investigate the limits at oo of each of the following functions defined on (0,00).
3x2 + 3x  1
x
*a. f(x)
b. f(x)
2r? + 1
1 + ?
4x + 1
2x + 3
*c. f(x)
Vx+1
*e. f(x) = V2+xx
1. f(x)
2Vx+
*&. f(x) = x cos
h. f(x) = x
x sin
18. Let f:(a,00) +R be such that lim xf(x) = L where LE R. Prove that lim f(x) = 0.
DEFINITION Lei E be a subset of R and f a realvalued function with domain E.
The function f is continuous at a point p E E, if for every e > 0, there exists a 8 > 0
such that
\f(x) – f(p)! 0, there exists a 8 > 0 such that
f(x) E N((p)) for all x € No(p) n E.
This is illustrated in Figure 4.7.
d. f(x)
=
X
Remarks
Vx 2x
=
+ 3x
(a) If p E E is a limit point of E, then f is continuous at p if and only if
lim f(x) = f(p).
rp
+00
100
1. Cauchy Course d'Analyse, p. 43.
4.2 Continuous Functions
141
142
Chapter 4
Limits and Continuity
Proof. That y is unique is clear. Let f(x) = x", which by Example 4.2.5(a) is contin
uous on R. Let a = 0 and b = y + 1. Since (y + 1)" > y.f satisfies the hypothesis of
Theorem 4.2.11. Thus there exists y, 0 < y < y + 1, such that
f(y) = y = y. O
4.2.14
COROLLARY If f:[0, 1] → [0, 1) is continuous, then there exists y € (0, 1) such
that f(y) = y.
Proof. Let g(x) = f(x) – x. Then 8(0) = f(0) = 0 and g(1) = f(1)  1 5 0. Thus
there exists y € (0, 1) such that 8(y) = 0; i.e., fly) = y. O
x
4.2.15
EXAMPLES
(a) In the proof of Theorem 4.2.11 continuity of the function f was required. The fol
lowing example shows that the converse of Theorem 4.2.11 is false; that is, if a func
tion f satisfies the intermediate value property on an interval (a, b), this does not im
ply that f is continuous on (a, b). Let f be defined on (0.7) as follows:
2
sin 0<
f(x) =
1,
x = 0.
Then f(0) = 1,5(%) = 1, and for every y, 1
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