Calculus I, Limits
1 / 36
Formal vs Informal Definition of a Limit
2 / 36
Formal vs Informal Definition of a Limit
Definition (Informal definition)
The limit of a function f (x), as x approaches a ∈ R, is L ∈ R, and we write
lim f (x) = L
x→a
if the values of f (x) can be made arbitrarily close to L by choosing the values
of x close enough to a.
2 / 36
Formal vs Informal Definition of a Limit
Definition (Informal definition)
The limit of a function f (x), as x approaches a ∈ R, is L ∈ R, and we write
lim f (x) = L
x→a
if the values of f (x) can be made arbitrarily close to L by choosing the values
of x close enough to a.
Definition (Formal definition)
We write
lim f (x) = L
x→a
if for any ε > 0 there exists δ > 0 such that
|x − a| < δ implies |f (x) − L| < ε.
2 / 36
Computing limits by definition
3 / 36
Computing limits by definition
To prove that
lim f (x) = L
x→a
consider a game in which your opponent makes a move by giving you a number
ε > 0 and you respond by producing a number δ > 0 (which depends on ε)
such that
|x − a| < δ implies |f (x) − L| < ε
3 / 36
Computing limits by definition
To prove that
lim f (x) = L
x→a
consider a game in which your opponent makes a move by giving you a number
ε > 0 and you respond by producing a number δ > 0 (which depends on ε)
such that
|x − a| < δ implies |f (x) − L| < ε
Example
Prove that
lim 1 = 1
x→1
3 / 36
Computing limits by definition
To prove that
lim f (x) = L
x→a
consider a game in which your opponent makes a move by giving you a number
ε > 0 and you respond by producing a number δ > 0 (which depends on ε)
such that
|x − a| < δ implies |f (x) − L| < ε
Example
Prove that
lim 1 = 1
x→1
Solution. We have f (x) = 1 for every x.
3 / 36
Computing limits by definition
To prove that
lim f (x) = L
x→a
consider a game in which your opponent makes a move by giving you a number
ε > 0 and you respond by producing a number δ > 0 (which depends on ε)
such that
|x − a| < δ implies |f (x) − L| < ε
Example
Prove that
lim 1 = 1
x→1
Solution. We have f (x) = 1 for every x. Suppose ε > 0 is given and we have
to find δ > 0 such that |x − 1| < δ implies |1 − 1| < ε.
3 / 36
Computing limits by definition
To prove that
lim f (x) = L
x→a
consider a game in which your opponent makes a move by giving you a number
ε > 0 and you respond by producing a number δ > 0 (which depends on ε)
such that
|x − a| < δ implies |f (x) − L| < ε
Example
Prove that
lim 1 = 1
x→1
Solution. We have f (x) = 1 for every x. Suppose ε > 0 is given and we have
to find δ > 0 such that |x − 1| < δ implies |1 − 1| < ε.
In this case any positive number δ works since |1 − 1| = 0 < ε. So, choose
δ = 1 - this is our response to any given ε.
3 / 36
Computing limits by definition
4 / 36
Computing limits by definition
Example
Prove that
lim x = 1
x→1
4 / 36
Computing limits by definition
Example
Prove that
lim x = 1
x→1
Example
Prove that
lim x 3 = 0
x→0
4 / 36
One-sided limits
5 / 36
One-sided limits
Similarly, one can defined one-sided limits as follows
5 / 36
One-sided limits
Similarly, one can defined one-sided limits as follows
Definition
We write
lim f (x) = L
x→a−
if the values of f (x) can be made arbitrarily close to L by choosing the values
of x close enough to a and less than a.
5 / 36
One-sided limits
Similarly, one can defined one-sided limits as follows
Definition
We write
lim f (x) = L
x→a−
if the values of f (x) can be made arbitrarily close to L by choosing the values
of x close enough to a and less than a.
We write
lim f (x) = L
x→a+
if the values of f (x) can be made arbitrarily close to L by choosing the values
of x close enough to a and greater than a.
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One-sided limits
6 / 36
One-sided limits
Fact
lim f (x) = L
x→a
6 / 36
One-sided limits
Fact
lim f (x) = L
x→a
if and only if
6 / 36
One-sided limits
Fact
lim f (x) = L
x→a
if and only if
lim f (x) = L and
x→a−
lim f (x) = L.
x→a+
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One-sided limits
Fact
lim f (x) = L
x→a
if and only if
lim f (x) = L and
x→a−
lim f (x) = L.
x→a+
Example
lim
x→0
|x|
=
x
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One-sided limits
Fact
lim f (x) = L
x→a
if and only if
lim f (x) = L and
x→a−
lim f (x) = L.
x→a+
Example
lim
x→0
|x|
= undefined
x
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Basic techniques
7 / 36
Basic techniques
Fact
The limit of a sum, difference, or product is equal to the sum, difference, or
product of the limits provided all the limits involved exist.
7 / 36
Basic techniques
Fact
The limit of a sum, difference, or product is equal to the sum, difference, or
product of the limits provided all the limits involved exist. The limit of a
quotient is the quotient of the limits provided that the limit of the denominator
is not 0.
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Basic techniques
Fact
The limit of a sum, difference, or product is equal to the sum, difference, or
product of the limits provided all the limits involved exist. The limit of a
quotient is the quotient of the limits provided that the limit of the denominator
is not 0.
Example (Positive example)
lim
x→0
x 3 + 2x 2 − 1
=
1 − 3x
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Basic techniques
Fact
The limit of a sum, difference, or product is equal to the sum, difference, or
product of the limits provided all the limits involved exist. The limit of a
quotient is the quotient of the limits provided that the limit of the denominator
is not 0.
Example (Positive example)
lim
x→0
x 3 + 2x 2 − 1
= −1
1 − 3x
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Basic techniques
Fact
The limit of a sum, difference, or product is equal to the sum, difference, or
product of the limits provided all the limits involved exist. The limit of a
quotient is the quotient of the limits provided that the limit of the denominator
is not 0.
Example (Positive example)
lim
x→0
x 3 + 2x 2 − 1
= −1
1 − 3x
Example (Negative example)
lim
x→2
x2 − 4
=
x −2
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Basic techniques
Fact
The limit of a sum, difference, or product is equal to the sum, difference, or
product of the limits provided all the limits involved exist. The limit of a
quotient is the quotient of the limits provided that the limit of the denominator
is not 0.
Example (Positive example)
lim
x→0
x 3 + 2x 2 − 1
= −1
1 − 3x
Example (Negative example)
lim
x→2
x2 − 4
=4
x −2
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Basic techniques
8 / 36
Basic techniques
Fact
If f (x) = g (x) for every x near a (but not equal to a) then
lim f (x) = lim g (x)
x→a
x→a
provided both exist.
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Basic techniques
Fact
If f (x) = g (x) for every x near a (but not equal to a) then
lim f (x) = lim g (x)
x→a
x→a
provided both exist.
2
−4
So, technically, in the previous example the function f (x) = xx−2
is being
replaced with the function g (x) = x + 2 whose limit is easy to compute as
x → 2.
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Basic techniques
Fact
If f (x) = g (x) for every x near a (but not equal to a) then
lim f (x) = lim g (x)
x→a
x→a
provided both exist.
2
−4
So, technically, in the previous example the function f (x) = xx−2
is being
replaced with the function g (x) = x + 2 whose limit is easy to compute as
x → 2.
Example
lim
x→0
(x + 2)2 − 4
=
x
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Basic techniques
Fact
If f (x) = g (x) for every x near a (but not equal to a) then
lim f (x) = lim g (x)
x→a
x→a
provided both exist.
2
−4
So, technically, in the previous example the function f (x) = xx−2
is being
replaced with the function g (x) = x + 2 whose limit is easy to compute as
x → 2.
Example
lim
x→0
(x + 2)2 − 4
=4
x
8 / 36
Basic techniques
9 / 36
Basic techniques
Example
lim
x→0
√
x2 + 4 − 2
=
x2
9 / 36
Basic techniques
Example
lim
x→0
√
x2 + 4 − 2
1
=
x2
4
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Basic techniques
Example
√
lim
x→0
x2 + 4 − 2
1
=
x2
4
Example
Compute limx→0 f (x) if
f (x) =
√
x +2
2 − x2
if x > 0,
if x < 0
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Basic techniques
Example
√
lim
x→0
x2 + 4 − 2
1
=
x2
4
Example
Compute limx→0 f (x) if
f (x) =
√
x +2
2 − x2
if x > 0,
if x < 0
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Basic techniques
10 / 36
Basic techniques
Fact
If f (x) 6 g (x) for every x near a (except possibly at a) and the limits of f (x)
and g (x) both exist as x → a then
lim f (x) 6 lim g (x)
x→a
x→a
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Basic techniques
Fact
If f (x) 6 g (x) for every x near a (except possibly at a) and the limits of f (x)
and g (x) both exist as x → a then
lim f (x) 6 lim g (x)
x→a
x→a
Fact (Squeeze Theorem)
If f (x) 6 g (x) 6 h(x) for every x near a (except possibly at a) and
limx→a f (x) = limx→a h(x) = L then
lim g (x) = L
x→a
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Basic techniques
Fact
If f (x) 6 g (x) for every x near a (except possibly at a) and the limits of f (x)
and g (x) both exist as x → a then
lim f (x) 6 lim g (x)
x→a
x→a
Fact (Squeeze Theorem)
If f (x) 6 g (x) 6 h(x) for every x near a (except possibly at a) and
limx→a f (x) = limx→a h(x) = L then
lim g (x) = L
x→a
Example
Show that limx→0 x 2 sin
1
x
= 0.
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Infinite limits
11 / 36
Infinite limits
Definition
We write limx→a f (x) = ∞ if the values of f (x) can be made arbitrarily large
by choosing the values of x close enough to a.
11 / 36
Infinite limits
Definition
We write limx→a f (x) = ∞ if the values of f (x) can be made arbitrarily large
by choosing the values of x close enough to a.
We write limx→a f (x) = −∞ if the values of f (x) can be made arbitrarily large
negative by choosing the values of x close enough to a.
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Infinite limits
Definition
We write limx→a f (x) = ∞ if the values of f (x) can be made arbitrarily large
by choosing the values of x close enough to a.
We write limx→a f (x) = −∞ if the values of f (x) can be made arbitrarily large
negative by choosing the values of x close enough to a.
One-sided infinite limits are defined similarly.
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Infinite limits
Definition
We write limx→a f (x) = ∞ if the values of f (x) can be made arbitrarily large
by choosing the values of x close enough to a.
We write limx→a f (x) = −∞ if the values of f (x) can be made arbitrarily large
negative by choosing the values of x close enough to a.
One-sided infinite limits are defined similarly.
Definition
x = a is called a vertical asymptote for f (x) if at least one one-sided limit of
f (x) as x approaches a is equal to ±∞.
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Infinite limits
12 / 36
Infinite limits
Example
lim
x→1−
1
=
x −1
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Infinite limits
Example
lim
x→1−
1
= −∞
x −1
12 / 36
Infinite limits
Example
lim
x→1−
1
= −∞
x −1
lim
x→1+
1
=
x −1
12 / 36
Infinite limits
Example
lim
x→1−
1
= −∞
x −1
lim
x→1+
1
=∞
x −1
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Infinite limits
Example
lim
x→1−
1
= −∞
x −1
lim
x→1+
1
=∞
x −1
Example
lim ln x =
x→0+
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Infinite limits
Example
lim
x→1−
1
= −∞
x −1
lim
x→1+
1
=∞
x −1
Example
lim ln x = − ∞
x→0+
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Infinite limits
Example
lim
x→1−
1
= −∞
x −1
lim
x→1+
1
=∞
x −1
Example
lim ln x = − ∞
x→0+
Example
lim tan x =
−
x→ π
2
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Infinite limits
Example
lim
x→1−
1
= −∞
x −1
lim
x→1+
1
=∞
x −1
Example
lim ln x = − ∞
x→0+
Example
lim tan x = ∞
−
x→ π
2
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Limits at infinity
13 / 36
Limits at infinity
Definition
We write
lim f (x) = L
x→∞
if the values of f (x) can be made arbitrarily close to L by choosing the values
of x large enough.
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Limits at infinity
Definition
We write
lim f (x) = L
x→∞
if the values of f (x) can be made arbitrarily close to L by choosing the values
of x large enough.
We write
lim f (x) = L
x→−∞
if the values of f (x) can be made arbitrarily close to L by choosing the values
of x large enough negative.
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Limits at infinity
Definition
We write
lim f (x) = L
x→∞
if the values of f (x) can be made arbitrarily close to L by choosing the values
of x large enough.
We write
lim f (x) = L
x→−∞
if the values of f (x) can be made arbitrarily close to L by choosing the values
of x large enough negative.
In either case we say that the horizontal line y = L is a horizontal asymptote
for f (x).
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Limits at infinity
14 / 36
Limits at infinity
Fact
If r > 0 then
lim
x→∞
1
= 0,
xr
14 / 36
Limits at infinity
Fact
If r > 0 then
lim
x→∞
1
= 0,
xr
lim
x→−∞
1
=0
xr
14 / 36
Limits at infinity
Fact
If r > 0 then
lim
x→∞
1
= 0,
xr
lim
x→−∞
1
=0
xr
Example
lim
x→∞
x2 − 1
=
x2 + 1
14 / 36
Limits at infinity
Fact
If r > 0 then
lim
x→∞
1
= 0,
xr
lim
x→−∞
1
=0
xr
Example
lim
x→∞
x2 − 1
=1
x2 + 1
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Limits at infinity
15 / 36
Limits at infinity
Example
lim √
x→∞
x +2
=
9x 2 + 1
15 / 36
Limits at infinity
Example
lim √
x→∞
x +2
1
=
3
9x 2 + 1
15 / 36
Limits at infinity
Example
lim √
x→∞
x +2
1
=
3
9x 2 + 1
Example
lim √
x→−∞
x +2
=
9x 2 + 1
15 / 36
Limits at infinity
Example
lim √
x→∞
x +2
1
=
3
9x 2 + 1
Example
lim √
x→−∞
x +2
1
= −
3
9x 2 + 1
15 / 36
Limits at infinity
16 / 36
Limits at infinity
Example
lim e x =
x→−∞
16 / 36
Limits at infinity
Example
lim e x = 0
x→−∞
16 / 36
Limits at infinity
Example
lim e x = 0
x→−∞
Example
lim
x→∞
x2
=
ex
16 / 36
Limits at infinity
Example
lim e x = 0
x→−∞
Example
lim
x→∞
x2
=0
ex
16 / 36
Limits at infinity
Example
lim e x = 0
x→−∞
Example
lim
x→∞
x2
=0
ex
Example
lim sin x =
x→∞
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Limits at infinity
Example
lim e x = 0
x→−∞
Example
lim
x→∞
x2
=0
ex
Example
lim sin x = undefined
x→∞
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Continuity at a point
17 / 36
Continuity at a point
Definition
A function f (x) is continuous at a ∈ R if
lim f (x) = f (a)
x→a
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Continuity at a point
Definition
A function f (x) is continuous at a ∈ R if
lim f (x) = f (a)
x→a
Remark
Continuity of f (x) at a implies that
17 / 36
Continuity at a point
Definition
A function f (x) is continuous at a ∈ R if
lim f (x) = f (a)
x→a
Remark
Continuity of f (x) at a implies that f (a) is defined (a ∈ Dom(f )),
17 / 36
Continuity at a point
Definition
A function f (x) is continuous at a ∈ R if
lim f (x) = f (a)
x→a
Remark
Continuity of f (x) at a implies that f (a) is defined (a ∈ Dom(f )), limx→a f (x)
exists,
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Continuity at a point
Definition
A function f (x) is continuous at a ∈ R if
lim f (x) = f (a)
x→a
Remark
Continuity of f (x) at a implies that f (a) is defined (a ∈ Dom(f )), limx→a f (x)
exists, and the equality from the definition holds.
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Continuity at a point
Definition
A function f (x) is continuous at a ∈ R if
lim f (x) = f (a)
x→a
Remark
Continuity of f (x) at a implies that f (a) is defined (a ∈ Dom(f )), limx→a f (x)
exists, and the equality from the definition holds.
Definition
If f (x) is not continuous at a then it is called discontinuous at a. There are
three types of discontinuity.
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Types of discontinuity
18 / 36
Types of discontinuity
Removable discontinuity:
18 / 36
Types of discontinuity
Removable discontinuity:
if a ∈
/ Dom(f ) but it is possible to extend f to a new function f
18 / 36
Types of discontinuity
Removable discontinuity:
if a ∈
/ Dom(f ) but it is possible to extend f to a new function f such that
f (x),
if x ∈ Dom(f ),
f (x) =
limx→a f (x), if x = a.
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Types of discontinuity
Removable discontinuity:
if a ∈
/ Dom(f ) but it is possible to extend f to a new function f such that
f (x),
if x ∈ Dom(f ),
f (x) =
limx→a f (x), if x = a.
Example
Check continuity of the function
f (x) =
x2 − 1
x −1
at the point x = 1.
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Types of discontinuity
19 / 36
Types of discontinuity
Infinite discontinuity:
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Types of discontinuity
Infinite discontinuity:
one of the limits limx→a− f (x), limx→a+ f (x) is equal to ±∞.
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Types of discontinuity
Infinite discontinuity:
one of the limits limx→a− f (x), limx→a+ f (x) is equal to ±∞.
Example
Check continuity of the function
f (x) =
1
x
at the point x = 0.
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Types of discontinuity
20 / 36
Types of discontinuity
Jump discontinuity:
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Types of discontinuity
Jump discontinuity:
both limx→a− f (x) and limx→a+ f (x) exist but are not equal to each other.
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Types of discontinuity
Jump discontinuity:
both limx→a− f (x) and limx→a+ f (x) exist but are not equal to each other.
Example
Check continuity of the function
f (x) =
|x|
x
at the point x = 0.
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Continuous functions
21 / 36
Continuous functions
Definition
f (x) is continuous on an interval if it is continuous at every point of the
interval.
21 / 36
Continuous functions
Definition
f (x) is continuous on an interval if it is continuous at every point of the
interval.
Fact
The following functions are continuous on their domains: polynomial, rational,
root, trigonometric, inverse trigonometric, exponential, logarithmic functions.
21 / 36
Continuous functions
Definition
f (x) is continuous on an interval if it is continuous at every point of the
interval.
Fact
The following functions are continuous on their domains: polynomial, rational,
root, trigonometric, inverse trigonometric, exponential, logarithmic functions.
Fact
Continuous functions (at a point, or on a given interval) are closed under taking
a sum, difference, product, and quotient (whenever the denominator is defined).
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Continuous functions
22 / 36
Continuous functions
Example
Where is the function continuous?
(a)
f (x) =
ln x + e x
x2 − 1
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Continuous functions
Example
Where is the function continuous?
(a)
f (x) =
ln x + e x
x2 − 1
(b)
f (x) =

 x +1

1
x
√
x −3
if x 6 1,
if 1 < x < 3,
if x > 3,
22 / 36
Continuous functions
Example
Where is the function continuous?
(a)
f (x) =
ln x + e x
x2 − 1
(b)
f (x) =

 x +1

1
x
√
x −3
if x 6 1,
if 1 < x < 3,
if x > 3,
Example
Find the values of the parameter c which make the function continuous
everywhere
cx 2 + 2x if x < 2,
f (x) =
x 3 − cx
if x > 2,
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Continuous functions
23 / 36
Continuous functions
Fact (Composition)
If f (x) is continuous at b and limx→a g (x) = b, then
23 / 36
Continuous functions
Fact (Composition)
If f (x) is continuous at b and limx→a g (x) = b, then
lim f (g (x)) = f lim g (x)
x→a
x→a
23 / 36
Continuous functions
Fact (Composition)
If f (x) is continuous at b and limx→a g (x) = b, then
lim f (g (x)) = f lim g (x)
x→a
x→a
Example
lim tan−1
x→2
x2 − 4
3x 2 − 6x
=
23 / 36
Continuous functions
Fact (Composition)
If f (x) is continuous at b and limx→a g (x) = b, then
lim f (g (x)) = f lim g (x)
x→a
x→a
Example
lim tan−1
x→2
x2 − 4
3x 2 − 6x
= tan−1
2
3
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Intermediate Value Theorem
24 / 36
Intermediate Value Theorem
Theorem
If f (x) is continuous on the closed interval [a, b] and N ∈ R is between f (a)
and f (b), where f (a) 6= f (b), then there exists c ∈ [a, b] such that f (c) = N.
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Intermediate Value Theorem
Theorem
If f (x) is continuous on the closed interval [a, b] and N ∈ R is between f (a)
and f (b), where f (a) 6= f (b), then there exists c ∈ [a, b] such that f (c) = N.
Example
Show that there is a root of the equation x 4 + x − 3 = 0 on the interval (1, 2).
24 / 36