1. Continuity

2. Continuity
The language of limits enables us to use algebra to
define and describe geometry such as “slopes
(derivatives)” of graphs of functions clearly.

3. Continuity
The language of limits enables us to use algebra to
define and describe geometry such as “slopes
(derivatives)” of graphs of functions clearly.
Let’s make some basic observations about graphs of
functions y = f(x).

4. Continuity
The language of limits enables us to use algebra to
define and describe geometry such as “slopes
(derivatives)” of graphs of functions clearly.
Let’s make some basic observations about graphs of
functions y = f(x). The graph of a function may be
an unbroken line or it may be broken at many places.

5. Continuity
The language of limits enables us to use algebra to
define and describe geometry such as “slopes
(derivatives)” of graphs of functions clearly.
Let’s make some basic observations about graphs of
functions y = f(x). The graph of a function may be
an unbroken line or it may be broken at many places.
If the curve is unbroken within a specified domain D,
we say the graph is continuous in D.

6. Continuity
The language of limits enables us to use algebra to
define and describe geometry such as “slopes
(derivatives)” of graphs of functions clearly.
Let’s make some basic observations about graphs of
functions y = f(x). The graph of a function may be
an unbroken line or it may be broken at many places.
If the curve is unbroken within a specified domain D,
we say the graph is continuous in D. (We will make
this notion more precise later.)

7. Continuity
The language of limits enables us to use algebra to
define and describe geometry such as “slopes
(derivatives)” of graphs of functions clearly.
Let’s make some basic observations about graphs of
functions y = f(x). The graph of a function may be
an unbroken line or it may be broken at many places.
If the curve is unbroken within a specified domain D,
we say the graph is continuous in D. (We will make
this notion more precise later.)
If the curve is broken within the domain D, we say
the graph is discontinuous in D.

8. Continuity
The language of limits enables us to use algebra to
define and describe geometry such as “slopes
(derivatives)” of graphs of functions clearly.
Let’s make some basic observations about graphs of
functions y = f(x). The graph of a function may be
an unbroken line or it may be broken at many places.
If the curve is unbroken within a specified domain D,
we say the graph is continuous in D. (We will make
this notion more precise later.)
If the curve is broken within the domain D, we say
the graph is discontinuous in D. More precisely, if the
graph breaks or becomes disconnected at x = a then
we say the function is discontinuous at x = a.

9. Continuity
Let’s look at the example from the last section.

10. Continuity
Let’s look at the example from the last section.
f(x) =
1 if 0 < x
x
y
y = f(x)

11. Continuity
Let’s look at the example from the last section.
f(x) =
1 if 0 < x
x
y
y = f(x)

12. Continuity
Let’s look at the example from the last section.
f(x) =
y
1 if 0 < x
0 if 0 = x x
y = f(x)

13. Continuity
Let’s look at the example from the last section.
f(x) =
y
1 if 0 < x
0 if 0 = x x
y = f(x)
(0,0)

14. Continuity
Let’s look at the example from the last section.
f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
x
y
y = f(x)
(0,0)

15. Continuity
Let’s look at the example from the last section.
f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
x
y
y = f(x)
(0,0)
The domain of f is the set of all real numbers R.

16. Continuity
Let’s look at the example from the last section.
f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
x
y
y = f(x)
(0,0)
The domain of f is the set of all real numbers R.
The graph has a jump at x = 0 so we say that
f is not continuous over R.

17. Continuity
Let’s look at the example from the last section.
f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
x
y
y = f(x)
(0,0)
The domain of f is the set of all real numbers R.
The graph has a jump at x = 0 so we say that
f is not continuous over R. But f is continuous over
the set of 0 < x (positive numbers) since there is no
break in that stretch of the graph.

18. Continuity
Let’s look at the example from the last section.
f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
x
y
y = f(x)
(0,0)
The domain of f is the set of all real numbers R.
The graph has a jump at x = 0 so we say that
f is not continuous over R. But f is continuous over
the set of 0 < x (positive numbers) since there is no
break in that stretch of the graph. It’s easy to see
where a graph is discontinuous at x = 0 but how do
we find discontinuity without looking at the graph?

19. Continuity
Let’s look at the example from the last section.
f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
x
y
y = f(x)
(0,0)
The precise question is that “given a function y = f(x)
and a point x = a in its domain, how can we
determine whether f(x) is continuous at x = a or not?

20. Continuity
Let’s look at the example from the last section.
f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
x
y
y = f(x)
The precise question is that “given a function y = f(x)
and a point x = a in its domain, how can we
determine whether f(x) is continuous at x = a or not?
An ant is crawling on the graph
above toward x = 0, how will it find
out that the line is broken at x = 0?
(0,0)
Q:

21. Continuity
Let’s look at the example from the last section.
f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
x
y
y = f(x)
(0,0)
The precise question is that “given a function y = f(x)
and a point x = a in its domain, how can we
determine whether f(x) is continuous at x = a or not?
It will find out because it will “stumble”
at x = 0 since the point (0, 1) is not
there, as it would have expected.
Q: A:
An ant is crawling on the graph
above toward x = 0, how will it find
out that the line is broken at x = 0?

22. Continuity
Let’s look at the example from the last section.
f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
x
y
y = f(x)
(0,1)
(0,0)
The precise question is that “given a function y = f(x)
and a point x = a in its domain, how can we
determine whether f(x) is continuous at x = a or not?
How do we characterize this “stumble” algebraically?
It will find out because it will “stumble”
at x = 0 since the point (0, 1) is not
there, as it would have expected.

23. Continuity
Let’s look at the example from the last section.
f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
x
y
y = f(x)
(0,1)
(0,0)
The precise question is that “given a function y = f(x)
and a point x = a in its domain, how can we
determine whether f(x) is continuous at x = a or not?
How do we characterize this “stumble” algebraically?
We stumbled due to the missing point at (0, 1):

24. Continuity
Let’s look at the example from the last section.
f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
x
y
y = f(x)
(0,1)
(0,0)
The precise question is that “given a function y = f(x)
and a point x = a in its domain, how can we
determine whether f(x) is continuous at x = a or not?
How do we characterize this “stumble” algebraically?
We stumbled due to the missing point at (0, 1):
lim f(x) = 1 means we expect that f(0) = 1 or that (0, 1)
0+
to be on the graph.

25. Continuity
Let’s look at the example from the last section.
f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
x
y
y = f(x)
(0,1)
(0,0)
The precise question is that “given a function y = f(x)
and a point x = a in its domain, how can we
determine whether f(x) is continuous at x = a or not?
How do we characterize this “stumble” algebraically?
We stumbled due to the missing point at (0, 1):
lim f(x) = 1 means we expect that f(0) = 1 or that (0, 1)
0+
to be on the graph. But f(0) = 0 so the graph skipped
(0, 1) and jumps to (0, 0), causes a discontinuity.

26. Continuity
Let’s consider the following function which fills in
the point (1, 0).
x
y
y = g(x)
(0,1)
g(x) =
1 if 0 ≤ x
–1 if x < 0

27. Continuity
Let’s consider the following function which fills in
the point (1, 0).
We see that lim g(x) = 1 = g(0) but that
0+
lim g(x) = –1 ≠ g(0) = 1.
0 –
x
y
y = g(x)
(0,1)
g(x) =
1 if 0 ≤ x
–1 if x < 0

28. Continuity
Let’s consider the following function which fills in
the point (1, 0).
We see that lim g(x) = 1 = g(0) but that
0+
lim g(x) = –1 ≠ g(0) = 1.
0 –
The ant coming from the right would not
stumble at x = 0 because
lim g(x) = 1 = g(0). But the ant coming
0+
from the left would find that there is a gap
because lim g(x) = –1 ≠ g(0) = 1.
0–
x
y
y = g(x)
(0,1)
g(x) =
1 if 0 ≤ x
–1 if x < 0

29. Continuity
Let’s consider the following function which fills in
the point (1, 0).
g(x) =
1 if 0 ≤ x
–1 if x < 0
x
y
y = g(x)
We see that lim g(x) = 1 = g(0) but that
0+
(0,1)
lim g(x) = –1 ≠ g(0) = 1. We say in this case that
0 –
g is right continuous but is not left continuous.

30. Continuity
Let’s consider the following function which fills in
the point (1, 0).
g(x) =
1 if 0 ≤ x
–1 if x < 0
x
y
y = g(x)
We see that lim g(x) = 1 = g(0) but that
0+
(0,1)
lim g(x) = –1 ≠ g(0) = 1. We say in this case that
0 –
g is right continuous but is not left continuous.
Definition. Given a function f and a point x = a.
We say f is right continuous at x = a if lim f(x) = f(a)
a+
and that f is not right continuous if lim f(x) ≠ f(a).
a+

31. Continuity
In a similar manner the function h(x) below is left
continuous at x = 0 but not right continuous at that
point.
h(x) =
1 if 0 < x
–1 if x ≤ 0
x
y
y = h(x)
(0,–1)

32. Continuity
In a similar manner the function h(x) below is left
continuous at x = 0 but not right continuous at that
point.
h(x) =
1 if 0 < x
–1 if x ≤ 0
x
y
y = h(x)
(0,–1)

33. Continuity
In a similar manner the function h(x) below is left
continuous at x = 0 but not right continuous at that
point.
h(x) =
1 if 0 < x
–1 if x ≤ 0
x
y
y = h(x)
(0,–1)
Definition. Given a function f and a point x = a in the
domain. We say f is left continuous if lim f(x) = f(a)
and that f is not left continuous if lim f(x) ≠ f(a).
a–
a–

34. Continuity
In a similar manner the function h(x) below is left
continuous at x = 0 but not right continuous at that
point.
h(x) =
1 if 0 < x
–1 if x ≤ 0
x
y
y = h(x)
(0,–1)
Definition. Given a function f and a point x = a in the
domain. We say f is left continuous if lim f(x) = f(a)
and that f is not left continuous if lim f(x) ≠ f(a).
a–
a–
Definition. Given a function f and a point x = a.
We say f is continuous if f is both right and left
continuous i.e. lim f(x) = lim f(x) = f(a).
a+ a–

35. Continuity
Here is a graphic summary.

36. Continuity
Here is a graphic summary.
x
y
y = g(x)
(0,1)
right but not left continuous at x =0

37. Continuity
Here is a graphic summary.
x
y
y = g(x)
(0,1)
right but not left continuous at x =0
x
y
y = h(x)
(0,–1)
left but not right continuous at x =0

38. Continuity
Here is a graphic summary.
x
y
y = g(x)
(0,1)
right but not left continuous at x =0
x
y
y = h(x)
(0,–1)
left but not right continuous at x =0
x
y
y = f(x)
(0,0)
neither right nor left continuous at x = 0

39. Continuity
Here is a graphic summary.
x
y
y = g(x)
(0,1)
right but not left continuous at x =0
x
y
y = h(x)
(0,–1)
left but not right continuous at x =0
x
y
y = f(x)
(0,0)
neither right nor left continuous at x = 0
x
y
y = 1
both left and right continuous
hence continuous at x =0

40. Continuity
Algebraic Steps for checking continuity

41. Continuity
Algebraic Steps for checking continuity
Given a function f and a point x = a,
calculate the right and the left limits as x goes to a±.

42. Continuity
Algebraic Steps for checking continuity
Given a function f and a point x = a,
calculate the right and the left limits as x goes to a±.
I. If either limit is UDF then f is discontinuous at x = a.

43. Continuity
Algebraic Steps for checking continuity
Given a function f and a point x = a,
calculate the right and the left limits as x goes to a±.
I. If either limit is UDF then f is discontinuous at x = a.
II. If both limits exist, let lim f(x) = R, lim f(x) = L,
a+ a–
compare R and L to the function output f(a).

44. Continuity
Algebraic Steps for checking continuity
Given a function f and a point x = a,
calculate the right and the left limits as x goes to a±.
I. If either limit is UDF then f is discontinuous at x = a.
II. If both limits exist, let lim f(x) = R, lim f(x) = L,
a+ a–
compare R and L to the function output f(a).
There are 4 possibilities.

45. Continuity
Algebraic Steps for checking continuity
Given a function f and a point x = a,
calculate the right and the left limits as x goes to a±.
I. If either limit is UDF then f is discontinuous at x = a.
II. If both limits exist, let lim f(x) = R, lim f(x) = L,
a+ a–
compare R and L to the function output f(a).
There are 4 possibilities.
• If R = L = f(a)
then f is cont.
at x = a

46. Continuity
Algebraic Steps for checking continuity
Given a function f and a point x = a,
calculate the right and the left limits as x goes to a±.
I. If either limit is UDF then f is discontinuous at x = a.
II. If both limits exist, let lim f(x) = R, lim f(x) = L,
a+ a–
compare R and L to the function output f(a).
There are 4 possibilities.
• If R = L = f(a)
then f is cont.
at x = a
• If R = f(a) ≠ L
then f is rt–cont.
but not left cont.
at x = a

47. Continuity
• If L = f(a) ≠ R
then f is left–cont.
but not rt. cont.
at x = a

48. Continuity
• If R ≠ L ≠ f(a)
then f is not rt.
nor left cont.
• If L = f(a) ≠ R
then f is left–cont.
but not rt. cont.
at x = a

49. Continuity
• If R ≠ L ≠ f(a)
then f is not rt.
nor left cont.
x
• If L = f(a) ≠ R
then f is left–cont.
but not rt. cont.
at x = a
Note that the function k(x) =
x
1
x ≠ 0
is precisely the constant x = 0
function y = 1,

50. Continuity
• If R ≠ L ≠ f(a)
then f is not rt.
nor left cont.
x
• If L = f(a) ≠ R
then f is left–cont.
but not rt. cont.
at x = a
Note that the function k(x) =
x
1
x ≠ 0
is precisely the constant x = 0
function y = 1, i.e. we fill in
the gap of x/x.

51. Continuity
• If R ≠ L ≠ f(a)
then f is not rt.
nor left cont.
x
• If L = f(a) ≠ R
then f is left–cont.
but not rt. cont.
at x = a
Note that the function k(x) =
x
1
x ≠ 0
is precisely the constant x = 0
function y = 1, i.e. we fill in
the gap of x/x.
x
y
(0, 1)

52. Continuity
• If R ≠ L ≠ f(a)
then f is not rt.
nor left cont.
x
• If L = f(a) ≠ R
then f is left–cont.
but not rt. cont.
at x = a
Note that the function k(x) =
x
1
x ≠ 0
is precisely the constant x = 0
function y = 1, i.e. we fill in
the gap of x/x.
x
y
(0, 1)

53. Continuity
• If R ≠ L ≠ f(a)
then f is not rt.
nor left cont.
x
• If L = f(a) ≠ R
then f is left–cont.
but not rt. cont.
at x = a
Note that the function k(x) =
x
1
x ≠ 0
is precisely the constant x = 0
function y = 1, i.e. we fill in
the gap of x/x. Such a
discontinuity that may be filled in
is said to be removable.
x
y
(0, 1)
A hole such as x=0 for x/x,
which may be filled, is said to
be a removable discontinuity.

54. Continuity
• If R ≠ L ≠ f(a)
then f is not rt.
nor left cont.
x
• If L = f(a) ≠ R
then f is left–cont.
but not rt. cont.
at x = a
Note that the function k(x) =
x
1
x ≠ 0
is precisely the constant x = 0
function y = 1, i.e. we fill in
the gap of x/x. Such a
discontinuity that may be filled in
is said to be removable.
For the above f(x), x = 0 is not
a removable discontinuity.
x
y
(0, 1)
A hole such as x=0 for x/x,
which may be filled, is said to
be a removable discontinuity.

55. Continuity
f(x) =
x2 – 4
x – 2
0 2 = x
sin(x – 2)
2 < x
x < 2
x – 2
Example A.
Is f(x) right or left cont.
at x = 2? Justify your
answers with limits.
Is it cont. at x = 2?

56. Continuity
f(x) =
x2 – 4
x – 2
0 2 = x
sin(x – 2)
2 < x
x < 2
x – 2
This is really x + 2
Example A.
Is f(x) right or left cont.
at x = 2? Justify your
answers with limits.
Is it cont. at x = 2?

57. Continuity
f(x) =
x – 2
0 2 = x
Calculate the right and the left limits lim f(x), lim f(x)
2–
x2 – 4
sin(x – 2)
2 < x
x < 2
x – 2
2+
This is really x + 2
Example A.
Is f(x) right or left cont.
at x = 2? Justify your
answers with limits.
Is it cont. at x = 2?

58. Continuity
f(x) =
x – 2
0 2 = x
Calculate the right and the left limits lim f(x), lim f(x)
The right limit of 2 is
2+
2–
x2 – 4
sin(x – 2)
2 < x
x < 2
x – 2
lim f(x) = x2 – 4
x – 2
lim
2+
2+
This is really x + 2
Example A.
Is f(x) right or left cont.
at x = 2? Justify your
answers with limits.
Is it cont. at x = 2?

59. Continuity
f(x) =
x – 2
0 2 = x
Calculate the right and the left limits lim f(x), lim f(x)
The right limit of 2 is
2+
2–
x2 – 4
sin(x – 2)
2 < x
x < 2
x – 2
lim f(x) = x2 – 4
lim lim
x – 2
=
(x + 2) = 4 = R
2+ 2+
2+
This is really x + 2
Example A.
Is f(x) right or left cont.
at x = 2? Justify your
answers with limits.
Is it cont. at x = 2?

60. Continuity
f(x) =
x – 2
0 2 = x
Calculate the right and the left limits lim f(x), lim f(x)
The right limit of 2 is
2+
2–
x2 – 4
sin(x – 2)
2 < x
x < 2
x – 2
lim f(x) = x2 – 4
lim lim
x – 2
=
(x + 2) = 4 = R
2+ 2+
2+
The left limit of 2 is
2– lim f(x) = lim
sin(x – 2)
2– x – 2
This is really x + 2
Example A.
Is f(x) right or left cont.
at x = 2? Justify your
answers with limits.
Is it cont. at x = 2?

61. Continuity
f(x) =
x – 2
0 2 = x
Calculate the right and the left limits lim f(x), lim f(x)
The right limit of 2 is
2+
2–
x2 – 4
sin(x – 2)
2 < x
x < 2
x – 2
lim f(x) = x2 – 4
lim lim
x – 2
=
(x + 2) = 4 = R
2+ 2+
2+
2– lim f(x) = lim 1 = L
=
The left limit of 2 is
sin(x – 2)
2– x – 2
This is really x + 2
Example A.
Is f(x) right or left cont.
at x = 2? Justify your
answers with limits.
Is it cont. at x = 2?

62. Continuity
f(x) =
x – 2
0 2 = x
Calculate the right and the left limits lim f(x), lim f(x)
The right limit of 2 is
2+
2–
x2 – 4
sin(x – 2)
2 < x
x < 2
x – 2
lim f(x) = x2 – 4
lim lim
x – 2
=
(x + 2) = 4 = R
2+ 2+
2+
2– lim f(x) = lim 1 = L
=
The left limit of 2 is
sin(x – 2)
x – 2
2–
The function value at x = 2 is f(2) = 0.
This is really x + 2
Example A.
Is f(x) right or left cont.
at x = 2? Justify your
answers with limits.
Is it cont. at x = 2?

63. Continuity
f(x) =
x – 2
This is really x + 2
0 2 = x
Example A.
Is f(x) right or left cont.
at x = 2? Justify your
answers with limits.
Is it cont. at x = 2?
Calculate the right and the left limits lim f(x), lim f(x)
The right limit of 2 is
2+
2–
x2 – 4
sin(x – 2)
2 < x
x < 2
x – 2
lim f(x) = x2 – 4
lim lim
x – 2
=
(x + 2) = 4 = R
2+ 2+
2+
2– lim f(x) = lim 1 = L
=
The left limit of 2 is
sin(x – 2)
x – 2
2–
The function value at x = 2 is f(2) = 0. Compare these
values

64. Continuity
f(x) =
x – 2
This is really x + 2
0 2 = x
Example A.
Is f(x) right or left cont.
at x = 2? Justify your
answers with limits.
Is it cont. at x = 2?
Calculate the right and the left limits lim f(x), lim f(x)
The right limit of 2 is
2+
2–
x2 – 4
sin(x – 2)
2 < x
x < 2
x – 2
lim f(x) = x2 – 4
lim lim
x – 2
=
(x + 2) = 4 = R
2+ 2+
2+
2– lim f(x) = lim 1 = L
=
The left limit of 2 is
sin(x – 2)
x – 2
2–
The function value at x = 2 is f(2) = 0. Compare these
values f(2) ≠ R ≠ L

65. Continuity
f(x) =
x – 2
This is really x + 2
0 2 = x
Example A.
Is f(x) right or left cont.
at x = 2? Justify your
answers with limits.
Is it cont. at x = 2?
Calculate the right and the left limits lim f(x), lim f(x)
The right limit of 2 is
2+
2–
x2 – 4
sin(x – 2)
2 < x
x < 2
x – 2
lim f(x) = x2 – 4
lim lim
x – 2
=
(x + 2) = 4 = R
2+ 2+
2+
2– lim f(x) = lim 1 = L
=
The left limit of 2 is
sin(x – 2)
x – 2
2–
The function value at x = 2 is f(2) = 0. Compare these
values f(2) ≠ R ≠ L so it’s not rt. nor left cont. at x= 2.

66. Continuity
Here is the graph of f(x).
f(x) =
x2 – 4
x – 2
(= x+2)
0 2 = x
sin(x – 2)
2 < x
x < 2
x – 2

67. Continuity
Here is the graph of f(x).
f(x) =
x2 – 4
x – 2
0 2 = x
sin(x – 2)
2 < x
x < 2
x – 2
x
y
(2,4)
y = x + 2
(= x+2)

68. Continuity
Here is the graph of f(x).
f(x) =
x2 – 4
x – 2
0 2 = x
sin(x – 2)
2 < x
x < 2
x – 2
y = x + 2
x
y
(2,4)
(2,f(2)=0)
(= x+2)

69. Continuity
Here is the graph of f(x).
f(x) =
x2 – 4
x – 2
0 2 = x
sin(x – 2)
2 < x
x < 2
x – 2
y = x + 2
x
y
(2,4)
(2,1)
(2,f(2)=0)
sin(x – 2)
x – 2
y =
(= x+2)

70. Continuity
Here is the graph of f(x).
f(x) =
x2 – 4
x – 2
0 2 = x
sin(x – 2)
2 < x
x < 2
x – 2
y = x + 2
x
y
(2,4)
(2,1)
(2,f(2)=0)
sin(x – 2)
x – 2
y =
(= x+2)
The discontinuity at x = 2 is not removable.

71. Continuity
Here is the graph of f(x).
f(x) =
x2 – 4
x – 2
0 2 = x
sin(x – 2)
2 < x
x < 2
x – 2
y = x + 2
x
y
(2,4)
(2,1)
(2,f(2)=0)
sin(x – 2)
x – 2
y =
(= x+2)
The discontinuity at x = 2 is not removable.
Note that if f is UDF at x = a then f is discontinuous
at x = a by default.

72. Continuity
Here is the graph of f(x).
f(x) =
x2 – 4
x – 2
0 2 = x
sin(x – 2)
2 < x
x < 2
x – 2
y = x + 2
x
y
(2,4)
(2,1)
(2,f(2)=0)
sin(x – 2)
x – 2
y =
(= x+2)
The discontinuity at x = 2 is not removable.
Note that if f is UDF at x = a then f is discontinuous
at x = a by default.
The point is that for y = x/x the discontinuity at x = 0
can be filled to make the graph continuous.

73. Continuity
Here is the graph of f(x).
f(x) =
x2 – 4
x – 2
0 2 = x
sin(x – 2)
2 < x
x < 2
x – 2
y = x + 2
x
y
(2,4)
(2,1)
(2,f(2)=0)
sin(x – 2)
x – 2
y =
(= x+2)
The discontinuity at x = 2 is not removable.
Note that if f is UDF at x = a then f is discontinuous
at x = a by default.
The point is that for y = x/x the discontinuity at x = 0
can be filled to make the graph continuous. But for
the f(x) above it’s not possible to fill the gap at x = 2.

74. Continuity
Example B.
sin(x2)
a. Is the discontinuity of y = x
at x = 0 removable?
Justify your answer.

75. Continuity
Example B.
sin(x2)
a. Is the discontinuity of y = x
at x = 0 removable?
Justify your answer.
The function is UDF hence discontinuous at x = 0.

76. Continuity
Example B.
sin(x2)
a. Is the discontinuity of y = x
at x = 0 removable?
Justify your answer.
The function is UDF hence discontinuous at x = 0.
Since lim sin(x2)
x
=
0

77. Continuity
Example B.
sin(x2)
a. Is the discontinuity of y = x
at x = 0 removable?
Justify your answer.
The function is UDF hence discontinuous at x = 0.
Since lim sin(x2)
x
= lim sin(x2)
x
x x
0 0

78. Continuity
Example B.
sin(x2)
a. Is the discontinuity of y = x
at x = 0 removable?
Justify your answer.
The function is UDF hence discontinuous at x = 0.
Since lim sin(x2)
x
= lim sin(x2)
x
x sin(x2)
= lim
x
x2
x
0 0 0

79. Continuity
Example B.
sin(x2)
a. Is the discontinuity of y = x
at x = 0 removable?
Justify your answer.
The function is UDF hence discontinuous at x = 0.
Since lim sin(x2)
x
= lim sin(x2)
x
x sin(x2)
= lim
x
x2
x
1
0 0 0

80. Continuity
Example B.
sin(x2)
a. Is the discontinuity of y = x
at x = 0 removable?
Justify your answer.
The function is UDF hence discontinuous at x = 0.
Since lim sin(x2)
x
= lim sin(x2)
x
x sin(x2)
= lim
x
x2
1
x = 0
0 0 0

81. Continuity
Example B.
sin(x2)
a. Is the discontinuity of y = x
at x = 0 removable?
Justify your answer.
The function is UDF hence discontinuous at x = 0.
Since lim sin(x2)
x
= lim sin(x2)
x
x sin(x2)
= lim
x
x2
1
x = 0
0 0 0
so that if we set y = 0 for x = 0
then y(0) = lim y.
0

82. Continuity
Example B.
sin(x2)
a. Is the discontinuity of y = x
at x = 0 removable?
Justify your answer.
The function is UDF hence discontinuous at x = 0.
Since lim sin(x2)
x
= lim sin(x2)
x
x sin(x2)
= lim
x
x2
1
x = 0
0 0 0
so that if we set y = 0 for x = 0
then y(0) = lim y.
0
This means (0,0) plugs the
gap and makes the function
continuous at x = 0.

83. Continuity
Example B.
sin(x2)
a. Is the discontinuity of y = x
at x = 0 removable?
Justify your answer.
The function is UDF hence discontinuous at x = 0.
Since lim sin(x2)
x
= lim sin(x2)
x
x sin(x2)
= lim
x
x2
1
x = 0
0 0 0
so that if we set y = 0 for x = 0
then y(0) = lim y.
0
This means (0,0) plugs the
gap and makes the function
continuous at x = 0.
Therefore x = 0 is a
removable discontinuity.

84. Continuity
Example B.
sin(x2)
a. Is the discontinuity of y = x
at x = 0 removable?
Justify your answer.
The function is UDF hence discontinuous at x = 0.
Since lim sin(x2)
x
= lim sin(x2)
x
x sin(x2)
= lim
x
x2
1
x = 0
0 0 0
so that if we set y = 0 for x = 0
then y(0) = lim y.
0
This means (0,0) plugs the
gap and makes the function
continuous at x = 0.
Therefore x = 0 is a
removable discontinuity.
y = sin(x2)
x is discontinuous at x = 0,
but it can be made continuous by
setting y = 0 at x = 0.

85. Continuity
sin(x)
b. Is the discontinuity of y = x2
at x = 0 removable?
Justify your answer.

86. Continuity
sin(x)
b. Is the discontinuity of y = x2
at x = 0 removable?
Justify your answer.
Calculate the right limit as x goes to 0+

87. Continuity
sin(x)
b. Is the discontinuity of y = x2
at x = 0 removable?
Justify your answer.
Calculate the right limit as x goes to 0+
lim sin(x)
x2 = lim sin(x)
1
x
0+ 0+ x

88. Continuity
sin(x)
b. Is the discontinuity of y = x2
at x = 0 removable?
Justify your answer.
Calculate the right limit as x goes to 0+
lim sin(x)
x2 = lim sin(x)
1
x
0+ 0+ x
1

89. Continuity
sin(x)
b. Is the discontinuity of y = x2
at x = 0 removable?
Justify your answer.
Calculate the right limit as x goes to 0+
lim sin(x)
x2 = lim sin(x)
1
x
= ∞
0+ 0+ x
1

90. Continuity
sin(x)
b. Is the discontinuity of y = x2
at x = 0 removable?
Justify your answer.
Calculate the right limit as x goes to 0+
lim sin(x)
x2 = lim sin(x)
1
x
= ∞
0+
0+ x
Hence the right limit does
not exist.

91. Continuity
sin(x)
b. Is the discontinuity of y = x2
at x = 0 removable?
Justify your answer.
Calculate the right limit as x goes to 0+
lim sin(x)
x2 = lim sin(x)
1
x
= ∞
0+
0+ x
Hence the right limit does
not exist. Therefore the
discontinuity at x = 0
is not removable.
1

92. Continuity
sin(x)
b. Is the discontinuity of y = x2
at x = 0 removable?
Justify your answer.
Calculate the right limit as x goes to 0+
lim sin(x)
x2 = lim sin(x)
1
x
= ∞
0+
0+ x
Hence the right limit does
not exist. Therefore the
discontinuity at x = 0
is not removable.
The graph of
is shown here.
1
y = sin(x)
x2
sin(x)
x2 y =

93. Continuity
Even though the discontinuity
is not removable at x = 0 for
sin(x)
y = x2
it’s clear that y goes to ±∞
as x got to 0.
sin(x)
x2 y =

94. Continuity
Even though the discontinuity
is not removable at x = 0 for
sin(x)
y = x2
it’s clear that y goes to ±∞
as x got to 0.
sin(x)
x2 y =
well behaved
discontinuity
at x = 0

95. Continuity
Even though the discontinuity
is not removable at x = 0 for
sin(x)
y = x2
it’s clear that y goes to ±∞
as x got to 0.
sin(x)
x2 y =
well behaved
discontinuity
at x = 0
For y = sin(1/x), it’s also UDF thus discontinuous
at x = 0.

96. Continuity
Even though the discontinuity
is not removable at x = 0 for
sin(x)
y = x2
it’s clear that y goes to ±∞
as x got to 0.
sin(x)
x2 y =
well behaved
discontinuity
at x = 0
For y = sin(1/x), it’s also UDF thus discontinuous
at x = 0. However it’s graph
is chaotic, swinging wildly
between 1 and –1. y = sin(1/x)
chaotic
discontinuity
at x= 0
1
–1

97. Continuity
Even though the discontinuity
is not removable at x = 0 for
sin(x)
y = x2
it’s clear that y goes to ±∞
as x got to 0.
sin(x)
x2 y =
There is no way for us to
estimate what y is even if we
know that x is small.
well behaved
discontinuity
at x = 0
For y = sin(1/x), it’s also UDF thus discontinuous
at x = 0. However it’s graph
is chaotic, swinging wildly
between 1 and –1. y = sin(1/x)
chaotic
discontinuity
at x= 0
1
–1