1. Section 2.4
Continuity
Math 1a
October 3, 2007

2. Questions
True or False
Right now there are two points on opposite sides of the Earth with
exactly the same temperature.

3. Questions
True or False
Right now there are two points on opposite sides of the Earth with
exactly the same temperature.
True or False
At one point in your life your height in inches equaled your weight
in pounds.

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

5. Direct Substitution Property
Theorem (The Direct Substitution Property)
If f is a polynomial or a rational function and a is in the domain of
f , then
lim f (x) = f (a)
x→a

7. Definition of Continuity
Definition
Let f be a function defined near a. We say that f is continuous at
a if
lim f (x) = f (a).
x→a

8. Free Theorems
Theorem
(a) Any polynomial is continuous everywhere; that is, it is
continuous on R = (−∞, ∞).
(b) Any rational function is continuous wherever it is defined; that
is, it is continuous on its domain.

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

11. Showing a function is continuous
Example
√
Let f (x) = 4x + 1. Show that f is continuous at 2.
Solution
We have
√
lim f (x) = lim 4x + 1
x→a x→2
= lim (4x + 1)
x→2
√
= 9 = 3.
Each step comes from the limit laws.

12. Showing a function is continuous
Example
√
Let f (x) = 4x + 1. Show that f is continuous at 2.
Solution
We have
√
lim f (x) = lim 4x + 1
x→a x→2
= lim (4x + 1)
x→2
√
= 9 = 3.
Each step comes from the limit laws.
In fact, f is continuous on its whole domain, which is − 1 , ∞ .
4

14. The Limit Laws give Continuity Laws
Theorem
If f and g are continuous at a and c is a constant, then the
following functions are also continuous at a:
1. f + g
2. f − g
3. cf
4. fg
f
5. (if g (a) = 0)
g

16. Transcendental functions are continuous, too
Theorem
The following functions are continuous wherever they are defined:
1. sin, cos, tan, cot sec, csc
2. x → ax , loga , ln
3. sin−1 , tan−1 , sec−1

18. What could go wrong?
In what ways could a function f fail to be continuous at a point a?
Look again at the definition:
lim f (x) = f (a)
x→a

19. Pitfall #1
Example
Let
x2 if 0 ≤ x ≤ 1
f (x) =
if 1 < x ≤ 2
2x
At which points is f continuous?

23. Pitfall #1: The limit does not exist
Example
Let
x2 if 0 ≤ x ≤ 1
f (x) =
if 1 < x ≤ 2
2x
At which points is f continuous?
Solution
At any point a in [0, 2] besides 1, lim f (x) = f (a) because f is
x→a
represented by a polynomial near a, and polynomials have the
direct substitution property. However,
lim f (x) = lim x 2 = 12 = 1
x→1− x→1−
lim f (x) = lim+ 2x = 2(1) = 2
x→1+ x→1
So f has no limit at 1. Therefore f is not continuous at 1.

25. Pitfall #2
Example
Let
x 2 + 2x + 1
f (x) =
x +1
At which points is f continuous?

27. Pitfall #2: The function has no value
Example
Let
x 2 + 2x + 1
f (x) =
x +1
At which points is f continuous?
Solution
Because f is rational, it is continuous on its whole domain. Note
that −1 is not in the domain of f , so f is not continuous there.

28. Pitfall #3
Example
Let
46 if x = 1
f (x) =
π if x = 1
At which points is f continuous?

29. Pitfall #3: function value = limit
Example
Let
46 if x = 1
f (x) =
π if x = 1
At which points is f continuous?
Solution
f is not continuous at 1 because f (1) = π but lim f (x) = 46.
x→1

30. Special types of discontinuites
removable discontinuity 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 discontinuity The limits lim f (x) and lim+ f (x) exist, but
x→a− x→a
are different. f (a) is one of these limits.

31. Special types of discontinuites
removable discontinuity 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 discontinuity The limits lim f (x) and lim+ f (x) exist, but
x→a− x→a
are different. f (a) is one of these limits.
The greatest integer function f (x) = [[x]] has jump discontinuities.

33. A Big Time Theorem
Theorem (The Intermediate Value Theorem)
Suppose that f is continuous 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.

34. Illustrating the IVT
f (x)
x

35. Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b]
f (x)
x

36. Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b]
f (x)
f (b)
f (a)
x
a b

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

38. Illustrating the IVT
Suppose that f is continuous 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.
f (x)
f (b)
N
f (a)
x
a c b

39. Illustrating the IVT
Suppose that f is continuous 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.
f (x)
f (b)
N
f (a)
x
a b

40. Illustrating the IVT
Suppose that f is continuous 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.
f (x)
f (b)
N
f (a)
x
a c1 c2 c3 b

41. Using the IVT
Example
Prove that the square root of two exists.

42. Using the IVT
Example
Prove that the square root of two exists.
Proof.
Let f (x) = x 2 , a continuous function on [1, 2].

43. Using the IVT
Example
Prove that the square root of two exists.
Proof.
Let f (x) = x 2 , a continuous function 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) = c 2 = 2.

44. Using the IVT
Example
Prove that the square root of two exists.
Proof.
Let f (x) = x 2 , a continuous function 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) = c 2 = 2.
In fact, we can “narrow in” on the square root of 2 by the method
of bisections.