1) The document provides an overview of continuity, including defining continuity as a function having a limit equal to its value at a point. 2) It discusses several theorems related to continuity, such as the sum of continuous functions being continuous and various trigonometric, exponential, and logarithmic functions being continuous on their domains. 3) The document also covers inverse trigonometric functions and their domains of continuity.

1. Sec on 1.5
Con nuity
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University
February 7, 2011
.

2. Announcements
Get-to-know-you extra
credit due Friday
February 11
Quiz 1 February 17/18 in
recita on

3. Objectives
Understand and apply the defini on of
con nuity for a func on at a point or
on an interval.
Given a piecewise defined func on,
decide where it is con nuous or
discon nuous.
State and understand the Intermediate
Value Theorem.
Use the IVT to show that a func on
takes a certain value, or that an
equa on has a solu on

4. Last time
Defini on
We write
lim f(x) = L
x→a
and say
“the limit of f(x), as x approaches a, equals L”
if we can make the values of f(x) arbitrarily close to L (as close to L
as we like) by taking x to be sufficiently close to a (on either side of
a) but not equal to a.

5. Basic Limits
Theorem (Basic Limits)
lim x = a
x→a
lim c = c
x→a

6. Limit Laws for arithmetic
Theorem (Limit Laws)
Let f and g be func ons with limits at a point a. Then
lim (f(x) + g(x)) = lim f(x) + lim g(x)
x→a x→a x→a
lim (f(x) − g(x)) = lim f(x) − lim g(x)
x→a x→a x→a
lim (f(x) · g(x)) = lim f(x) · lim g(x)
x→a x→a x→a
f(x) limx→a f(x)
lim = if lim g(x) ̸= 0
x→a g(x) limx→a g(x) x→a

7. Hatsumon
Here are some discussion ques ons to start.
True or False
At some point in your life you were exactly three feet tall.

8. Hatsumon
Here are some discussion ques ons to start.
True or False
At some point in your life you were exactly three feet tall.
True or False
At some point in your life your height (in inches) was equal to your
weight (in pounds).

9. Hatsumon
Here are some discussion ques ons to start.
True or False
At some point in your life you were exactly three feet tall.
True or False
At some point in your life your height (in inches) was equal to your
weight (in pounds).
True or False
Right now there are a pair of points on opposite sides of the world
measuring the exact same temperature.

10. Outline
Con nuity
The Intermediate Value Theorem
Back to the Ques ons

11. Recall: Direct Substitution
Property
Theorem (The Direct Subs tu on Property)
If f is a polynomial or a ra onal func on and a is in the domain of f,
then
lim f(x) = f(a)
x→a
This property is so useful it’s worth naming.

12. Definition of Continuity
y
Defini on
Let f be a func on defined near
a. We say that f is con nuous at f(a)
a if
lim f(x) = f(a).
x→a
. x
a

13. Definition of Continuity
y
Defini on
Let f be a func on defined near
a. We say that f is con nuous at f(a)
a if
lim f(x) = f(a).
x→a
A func on f is con nuous if it is
con nuous at every point in its
domain. . x
a

14. Scholium
Defini on
Let f be a func on defined near a. We say that f is con nuous at a if
lim f(x) = f(a).
x→a
There are three important parts to this defini on.
The func on has to have a limit at a,
the func on has to have a value at a,
and these values have to agree.

15. Free Theorems
Theorem
(a) Any polynomial is con nuous everywhere; that is, it is
con nuous on R = (−∞, ∞).
(b) Any ra onal func on is con nuous wherever it is defined; that is,
it is con nuous on its domain.

16. Showing a function is continuous
.
Example
√
Let f(x) = 4x + 1. Show that f is con nuous at 2.

17. Showing a function is continuous
.
Example
√
Let f(x) = 4x + 1. Show that f is con nuous at 2.
Solu on
We want to show that lim f(x) = f(2). We have
x→2
√ √ √
lim f(x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f(2).
x→a x→2 x→2
Each step comes from the limit laws.

18. At which other points?
Ques on
√
As before, let f(x) = 4x + 1. At which points is f con nuous?

19. At which other points?
Ques on
√
As before, let f(x) = 4x + 1. At which points is f con nuous?
Solu on
If a > −1/4, then lim (4x + 1) = 4a + 1 > 0, so
x→a
√ √ √
lim f(x) = lim 4x + 1 = lim (4x + 1) = 4a + 1 = f(a)
x→a x→a x→a
and f is con nuous at a.

20. At which other points?
Ques on
√
As before, let f(x) = 4x + 1. At which points is f con nuous?
Solu on
If a = −1/4, then 4x + 1 < 0 to the le of a, which means
√
4x + 1 is undefined. S ll,
√ √ √
lim+ f(x) = lim+ 4x + 1 = lim+ (4x + 1) = 0 = 0 = f(a)
x→a x→a x→a
so f is con nuous on the right at a = −1/4.

21. Limit Laws give Continuity Laws
Theorem
If f(x) and g(x) are con nuous at a and c is a constant, then the
following func ons are also con nuous at a:
(f + g)(x) (fg)(x)
(f − g)(x) f
(x) (if g(a) ̸= 0)
(cf)(x) g

22. Sum of continuous is continuous
We want to show that
lim (f + g)(x) = (f + g)(a).
x→a

23. Sum of continuous is continuous
We want to show that
lim (f + g)(x) = (f + g)(a).
x→a
We just follow our nose.

24. Sum of continuous is continuous
We want to show that
lim (f + g)(x) = (f + g)(a).
x→a
We just follow our nose.
(def of f + g) lim (f + g)(x)
x→a

25. Sum of continuous is continuous
We want to show that
lim (f + g)(x) = (f + g)(a).
x→a
We just follow our nose.
(def of f + g) lim (f + g)(x) = lim [f(x) + g(x)]
x→a x→a

26. Sum of continuous is continuous
We want to show that
lim (f + g)(x) = (f + g)(a).
x→a
We just follow our nose.
(def of f + g) lim (f + g)(x) = lim [f(x) + g(x)]
x→a x→a
(if these limits exist)

27. Sum of continuous is continuous
We want to show that
lim (f + g)(x) = (f + g)(a).
x→a
We just follow our nose.
(def of f + g) lim (f + g)(x) = lim [f(x) + g(x)]
x→a x→a
(if these limits exist) = lim f(x) + lim g(x)
x→a x→a

28. Sum of continuous is continuous
We want to show that
lim (f + g)(x) = (f + g)(a).
x→a
We just follow our nose.
(def of f + g) lim (f + g)(x) = lim [f(x) + g(x)]
x→a x→a
(if these limits exist) = lim f(x) + lim g(x)
x→a x→a
(they do; f and g are cts.)

29. Sum of continuous is continuous
We want to show that
lim (f + g)(x) = (f + g)(a).
x→a
We just follow our nose.
(def of f + g) lim (f + g)(x) = lim [f(x) + g(x)]
x→a x→a
(if these limits exist) = lim f(x) + lim g(x)
x→a x→a
(they do; f and g are cts.) = f(a) + g(a)

30. Sum of continuous is continuous
We want to show that
lim (f + g)(x) = (f + g)(a).
x→a
We just follow our nose.
(def of f + g) lim (f + g)(x) = lim [f(x) + g(x)]
x→a x→a
(if these limits exist) = lim f(x) + lim g(x)
x→a x→a
(they do; f and g are cts.) = f(a) + g(a)
(def of f + g again)

31. Sum of continuous is continuous
We want to show that
lim (f + g)(x) = (f + g)(a).
x→a
We just follow our nose.
(def of f + g) lim (f + g)(x) = lim [f(x) + g(x)]
x→a x→a
(if these limits exist) = lim f(x) + lim g(x)
x→a x→a
(they do; f and g are cts.) = f(a) + g(a)
(def of f + g again) = (f + g)(a)

32. Sum of continuous is continuous
We want to show that
lim (f + g)(x) = (f + g)(a).
x→a
We just follow our nose.
(def of f + g) lim (f + g)(x) = lim [f(x) + g(x)]
x→a x→a
(if these limits exist) = lim f(x) + lim g(x)
x→a x→a
(they do; f and g are cts.) = f(a) + g(a)
(def of f + g again) = (f + g)(a)

33. Trig functions are continuous
sin and cos are con nuous
on R.
.
sin

34. Trig functions are continuous
sin and cos are con nuous
on R.
cos
.
sin

35. Trig functions are continuous
tan
sin and cos are con nuous
on R.
sin 1
tan = and sec = are
cos cos cos
con nuous on their domain,
{π
which is } .
sin
R + kπ k ∈ Z .
2

36. Trig functions are continuous
tan sec
sin and cos are con nuous
on R.
sin 1
tan = and sec = are
cos cos cos
con nuous on their domain,
{π
which is } .
sin
R + kπ k ∈ Z .
2

37. Trig functions are continuous
tan sec
sin and cos are con nuous
on R.
sin 1
tan = and sec = are
cos cos cos
con nuous on their domain,
{π
which is } .
sin
R + kπ k ∈ Z .
2
cos 1
cot = and csc = are
sin sin
con nuous on their domain,
which is R { kπ | k ∈ Z }. cot

38. Trig functions are continuous
tan sec
sin and cos are con nuous
on R.
sin 1
tan = and sec = are
cos cos cos
con nuous on their domain,
{π
which is } .
sin
R + kπ k ∈ Z .
2
cos 1
cot = and csc = are
sin sin
con nuous on their domain,
which is R { kπ | k ∈ Z }. cot csc

39. Exp and Log are continuous
For any base a 1,
the func on x → ax is ax
con nuous on R
.

40. Exp and Log are continuous
For any base a 1,
the func on x → ax is ax
loga x
con nuous on R
the func on loga is
con nuous on its .
domain: (0, ∞)

41. Exp and Log are continuous
For any base a 1,
the func on x → ax is ax
loga x
con nuous on R
the func on loga is
con nuous on its .
domain: (0, ∞)
In par cular ex and
ln = loge are con nuous
on their domains

42. Inverse trigonometric functions
are mostly continuous
sin−1 and cos−1 are con nuous on (−1, 1), le con nuous at 1,
and right con nuous at −1.
π
π/2
−1 .
sin

43. Inverse trigonometric functions
are mostly continuous
sin−1 and cos−1 are con nuous on (−1, 1), le con nuous at 1,
and right con nuous at −1.
π
cos−1
π/2
−1 .
sin

44. Inverse trigonometric functions
are mostly continuous
sin−1 and cos−1 are con nuous on (−1, 1), le con nuous at 1,
and right con nuous at −1.
sec−1 and csc−1 are con nuous on (−∞, −1) ∪ (1, ∞), le
con nuous at −1, and right con nuous at 1.
π
cos−1 sec−1
π/2
−1 .
sin

45. Inverse trigonometric functions
are mostly continuous
sin−1 and cos−1 are con nuous on (−1, 1), le con nuous at 1,
and right con nuous at −1.
sec−1 and csc−1 are con nuous on (−∞, −1) ∪ (1, ∞), le
con nuous at −1, and right con nuous at 1.
π
cos−1 sec−1
π/2
−1
csc−1
sin .

46. Inverse trigonometric functions
are mostly continuous
sin−1 and cos−1 are con nuous on (−1, 1), le con nuous at 1,
and right con nuous at −1.
sec−1 and csc−1 are con nuous on (−∞, −1) ∪ (1, ∞), le
con nuous at −1, and right con nuous at 1.
tan−1 and cot−1 are con nuous on R.
π
cos−1 sec−1
π/2
−1
tan−1
csc−1
sin .

47. Inverse trigonometric functions
are mostly continuous
sin−1 and cos−1 are con nuous on (−1, 1), le con nuous at 1,
and right con nuous at −1.
sec−1 and csc−1 are con nuous on (−∞, −1) ∪ (1, ∞), le
con nuous at −1, and right con nuous at 1.
tan−1 and cot−1 are con nuous on R.
π
cot−1 cos−1 sec−1
π/2
−1
tan−1
csc−1
sin .

48. What could go wrong?
In what ways could a func on f fail to be con nuous at a point a?
Look again at the equa on from the defini on:
lim f(x) = f(a)
x→a

49. Continuity FAIL
.
Example
{
x2 if 0 ≤ x ≤ 1
Let f(x) = . At which points is f con nuous?
2x if 1 x ≤ 2

50. Continuity FAIL: no limit
.
Example
{
x2 if 0 ≤ x ≤ 1
Let f(x) = . At which points is f con nuous?
2x if 1 x ≤ 2
Solu on
At any point a besides 1, lim f(x) = f(a) because f is represented by a
x→a
polynomial near a, and polynomials have the direct subs tu on property.
lim− f(x) = lim− x2 = 12 = 1 and lim+ f(x) = lim+ 2x = 2(1) = 2
x→1 x→1 x→1 x→1
So f has no limit at 1. Therefore f is not con nuous at 1.

51. Graphical Illustration of Pitfall #1
y
4
3 The func on cannot be
con nuous at a point if the
2
func on has no limit at that
1 point.
. x
−1 1 2
−1

52. Graphical Illustration of Pitfall #1
y
4
3 The func on cannot be
con nuous at a point if the
2
1 FAIL func on has no limit at that
point.
. x
−1 1 2
−1

53. Continuity FAIL
Example
Let
x2 + 2x + 1
f(x) =
x+1
At which points is f con nuous?

54. Continuity FAIL: no value
Example
Let
x2 + 2x + 1
f(x) =
x+1
At which points is f con nuous?
Solu on
Because f is ra onal, it is con nuous on its whole domain. Note that
−1 is not in the domain of f, so f is not con nuous there.

55. Graphical Illustration of Pitfall #2
y
1 The func on cannot be
con nuous at a point outside
. x its domain (that is, a point
−1 where it has no value).

56. Graphical Illustration of Pitfall #2
y
FAIL
1 The func on cannot be
con nuous at a point outside
. x its domain (that is, a point
−1 where it has no value).

57. Continuity FAIL
Example
Let {
7 if x ̸= 1
f(x) =
π if x = 1
At which points is f con nuous?

58. Continuity FAIL: value ̸= limit
Example
Let {
7 if x ̸= 1
f(x) =
π if x = 1
At which points is f con nuous?
Solu on
f is not con nuous at 1 because f(1) = π but lim f(x) = 7.
x→1

59. Graphical Illustration of Pitfall #3
y
7 If the func on has a limit and
a value at a point the two
π must s ll agree.
. x
1

60. Graphical Illustration of Pitfall #3
y
7 If the func on has a limit and
a value at a point the two
FAIL
π
. x
must s ll agree.
1

61. Special types of discontinuities
removable discon nuity The limit lim f(x) exists, but f is not
x→a
defined at a or its value at a is not equal to the limit at a.
jump discon nuity The limits lim− f(x) and lim+ f(x) exist, but are
x→a x→a
different.

62. Special discontinuities graphically
y y
7 2
π 1
. x . x
1 1
removable jump

63. Special discontinuities graphically
y y
Presto! con nuous!
7 2
π 1
. x . x
1 1
removable jump

64. Special discontinuities graphically
y y
Presto! con nuous!
7 2
π 1 con nuous?
. x . x
1 1
removable jump

65. Special discontinuities graphically
y y
Presto! con nuous!
7 2 con nuous?
π 1
. x . x
1 1
removable jump

66. Special discontinuities graphically
y y
Presto! con nuous!
7 2
con nuous?
π 1
. x . x
1 1
removable jump

67. Special types of discontinuities
removable discon nuity The limit lim f(x) exists, but f is not
x→a
defined at a or its value at a is not equal to the limit at a.
By re-defining f(a) = lim f(x), f can be made con nuous
x→a
at a
jump discon nuity The limits lim− f(x) and lim+ f(x) exist, but are
x→a x→a
different.

68. Special types of discontinuities
removable discon nuity The limit lim f(x) exists, but f is not
x→a
defined at a or its value at a is not equal to the limit at a.
By re-defining f(a) = lim f(x), f can be made con nuous
x→a
at a
jump discon nuity The limits lim− f(x) and lim+ f(x) exist, but are
x→a x→a
different. The func on cannot be made con nuous by
changing a single value.

69. The greatest integer function
[[x]] is the greatest integer ≤ x. y
3
x [[x]] y = [[x]]
0 0 2
1 1
1.5 1 1
1.9 1 . x
2.1 2 −2 −1 1 2 3
−0.5 −1 −1
−0.9 −1
−1.1 −2 −2

70. The greatest integer function
[[x]] is the greatest integer ≤ x. y
3
x [[x]] y = [[x]]
0 0 2
1 1
1.5 1 1
1.9 1 . x
2.1 2 −2 −1 1 2 3
−0.5 −1 −1
−0.9 −1
−1.1 −2 −2
This func on has a jump discon nuity at each integer.

71. Outline
Con nuity
The Intermediate Value Theorem
Back to the Ques ons

72. A Big Time Theorem
Theorem (The Intermediate Value Theorem)
Suppose that f is con nuous on the closed interval [a, b] and let N be
any number between f(a) and f(b), where f(a) ̸= f(b). Then there
exists a number c in (a, b) such that f(c) = N.

73. Illustrating the IVT
f(x)
Theorem
. x

74. Illustrating the IVT
f(x)
Theorem
Suppose that f is con nuous
on the closed interval [a, b]
. x

75. Illustrating the IVT
f(x)
Theorem
Suppose that f is con nuous f(b)
on the closed interval [a, b]
f(a)
. a b x

76. Illustrating the IVT
f(x)
Theorem
Suppose that f is con nuous f(b)
on the closed interval [a, b]
N
and let N be any number
between f(a) and f(b), where f(a)
f(a) ̸= f(b).
. a b x

77. Illustrating the IVT
f(x)
Theorem
Suppose that f is con nuous f(b)
on the closed interval [a, b]
N
and let N be any number
between f(a) and f(b), where f(a)
f(a) ̸= f(b). Then there exists
a number c in (a, b) such that
f(c) = N. . a c b x

78. Illustrating the IVT
f(x)
Theorem
Suppose that f is con nuous f(b)
on the closed interval [a, b]
N
and let N be any number
between f(a) and f(b), where f(a)
f(a) ̸= f(b). Then there exists
a number c in (a, b) such that
f(c) = N. . a b x

79. Illustrating the IVT
f(x)
Theorem
Suppose that f is con nuous f(b)
on the closed interval [a, b]
N
and let N be any number
between f(a) and f(b), where f(a)
f(a) ̸= f(b). Then there exists
a number c in (a, b) such that
f(c) = N. . ac1 c2 c3b x

80. What the IVT does not say
The Intermediate Value Theorem is an “existence” theorem.
It does not say how many such c exist.
It also does not say how to find c.
S ll, it can be used in itera on or in conjunc on with other
theorems to answer these ques ons.

81. Using the IVT to find zeroes
Example
Let f(x) = x3 − x − 1. Show that there is a zero for f on the interval
[1, 2].

82. Using the IVT to find zeroes
Example
Let f(x) = x3 − x − 1. Show that there is a zero for f on the interval
[1, 2].
Solu on
f(1) = −1 and f(2) = 5. So there is a zero between 1 and 2.

83. Using the IVT to find zeroes
Example
Let f(x) = x3 − x − 1. Show that there is a zero for f on the interval
[1, 2].
Solu on
f(1) = −1 and f(2) = 5. So there is a zero between 1 and 2.
In fact, we can “narrow in” on the zero by the method of bisec ons.

84. Finding a zero by bisection
y
x f(x)
. x

85. Finding a zero by bisection
y
x f(x)
1 −1
. x

86. Finding a zero by bisection
y
x f(x)
1 −1
2 5
. x

87. Finding a zero by bisection
y
x f(x)
1 −1
1.5 0.875
2 5
. x

88. Finding a zero by bisection
y
x f(x)
1 −1
1.25 − 0.296875
1.5 0.875
2 5
. x

89. Finding a zero by bisection
y
x f(x)
1 −1
1.25 − 0.296875
1.375 0.224609
1.5 0.875
2 5
. x

90. Finding a zero by bisection
y
x f(x)
1 −1
1.25 − 0.296875
1.3125 − 0.0515137
1.375 0.224609
1.5 0.875
2 5
. x

91. Finding a zero by bisection
y
x f(x)
1 −1
1.25 − 0.296875
1.3125 − 0.0515137
1.375 0.224609
1.5 0.875
2 5
. x
(More careful analysis yields
1.32472.)

92. Using the IVT to assert existence
of numbers
Example
Suppose we are unaware of the square root func on and that it’s
con nuous. Prove that the square root of two exists.

93. Using the IVT to assert existence
of numbers
Example
Suppose we are unaware of the square root func on and that it’s
con nuous. Prove that the square root of two exists.
Proof.
Let f(x) = x2 , a con nuous func on on [1, 2].

94. Using the IVT to assert existence
of numbers
Example
Suppose we are unaware of the square root func on and that it’s
con nuous. Prove that the square root of two exists.
Proof.
Let f(x) = x2 , a con nuous func on on [1, 2]. Note f(1) = 1 and
f(2) = 4. Since 2 is between 1 and 4, there exists a point c in (1, 2)
such that
f(c) = c2 = 2.

95. Outline
Con nuity
The Intermediate Value Theorem
Back to the Ques ons

96. Back to the Questions
True or False
At one point in your life you were exactly three feet tall.

97. Question 1
Answer
The answer is TRUE.
Let h(t) be height, which varies con nuously over me.
Then h(birth) 3 ft and h(now) 3 ft.
So by the IVT there is a point c in (birth, now) where h(c) = 3.

98. Back to the Questions
True or False
At one point in your life you were exactly three feet tall.
True or False
At one point in your life your height in inches equaled your weight in
pounds.

99. Question 2
Answer
The answer is TRUE.
Let h(t) be height in inches and w(t) be weight in pounds, both
varying con nuously over me.
Let f(t) = h(t) − w(t).
For most of us (call your mom), f(birth) 0 and f(now) 0.
So by the IVT there is a point c in (birth, now) where f(c) = 0.
In other words,
h(c) − w(c) = 0 ⇐⇒ h(c) = w(c).

100. Back to the Questions
True or False
At one point in your life you were exactly three feet tall.
True or False
At one point in your life your height in inches equaled your weight in
pounds.
True or False
Right now there are two points on opposite sides of the Earth with
exactly the same temperature.

101. Question 3
Answer
The answer is TRUE.
Let T(θ) be the temperature at the point on the equator at
longitude θ.
How can you express the statement that the temperature on
opposite sides is the same?
How can you ensure this is true?

102. Question 3
Let f(θ) = T(θ) − T(θ + 180◦ )
Then
f(0) = T(0) − T(180)
while
f(180) = T(180) − T(360) = −f(0)
So somewhere between 0 and 180 there is a point θ where
f(θ) = 0!

103. Summary
What have we learned today?
Defini on: a func on is con nuous at a point if the limit of the
func on at that point agrees with the value of the func on at
that point.
We o en make a fundamental assump on that func ons we
meet in nature are con nuous.
The Intermediate Value Theorem is a basic property of real
numbers that we need and use a lot.