## Full text

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WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

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## 4.3 Mean Value Theorem

### Lemma:

[Rolle’s Theorem]

Suppose that a, b ∈R with a 6= b. If f is continuous on [a,b], differentiable on (a, b), and if f(a)=f(b), then f0(c) = 0 for some c ∈ (a, b).

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## 4.3 Mean Value Theorem

### Lemma:

[Rolle’s Theorem]

Suppose that a, b ∈R with a 6= b. If f is continuous on [a,b], differentiable on (a, b), and if f(a)=f(b), then f0(c) = 0 for some c ∈ (a, b).

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### Proof:

By the Extreme Value Theorem, f has a finite maximum M and a finite minimum m on [a,b]. If M = m, then f is

constant on (a,b) and f’(x)=0 for all x ∈ (a, b). Suppose that M 6= m. Since f (a) = f (b), f must assume one of the values M or m at some point c ∈ (a, b). By symmetry, we may suppose that f (c) = M. (That is, if we can prove the theorem when f (c) = M, then a similar proof establishes the theorem when f (c) = m.)

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### Proof:

By the Extreme Value Theorem,f has a finite maximum M and a finite minimum m on [a,b]. If M = m,then f is

constant on (a,b) and f’(x)=0 for all x ∈ (a, b). Suppose that M 6= m. Since f (a) = f (b), f must assume one of the values M or m at some point c ∈ (a, b). By symmetry, we may suppose that f (c) = M. (That is, if we can prove the theorem when f (c) = M, then a similar proof establishes the theorem when f (c) = m.)

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### Proof:

By the Extreme Value Theorem, f has a finite maximum M and a finite minimum m on [a,b]. If M = m, then f is

constant on (a,b) and f’(x)=0 for all x ∈ (a, b).Suppose that M 6= m. Since f (a) = f (b), f must assume one of the values M or m at some point c ∈ (a, b). By symmetry, we may suppose that f (c) = M. (That is, if we can prove the theorem when f (c) = M, then a similar proof establishes the theorem when f (c) = m.)

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### Proof:

By the Extreme Value Theorem, f has a finite maximum M and a finite minimum m on [a,b]. If M = m,then f is

constant on (a,b) and f’(x)=0 for all x ∈ (a, b). Suppose that M 6= m. Since f (a) = f (b), f must assume one of the values M or m at some point c ∈ (a, b). By symmetry, we may suppose that f (c) = M. (That is, if we can prove the theorem when f (c) = M, then a similar proof establishes the theorem when f (c) = m.)

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### Proof:

By the Extreme Value Theorem, f has a finite maximum M and a finite minimum m on [a,b]. If M = m, then f is

constant on (a,b) and f’(x)=0 for all x ∈ (a, b).Suppose that M 6= m. Since f (a) = f (b),f must assume one of the values M or m at some point c ∈ (a, b). By symmetry, we may suppose that f (c) = M. (That is, if we can prove the theorem when f (c) = M, then a similar proof establishes the theorem when f (c) = m.)

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### Proof:

By the Extreme Value Theorem, f has a finite maximum M and a finite minimum m on [a,b]. If M = m, then f is

constant on (a,b) and f’(x)=0 for all x ∈ (a, b). Suppose that M 6= m. Since f (a) = f (b), f must assume one of the values M or m at some point c ∈ (a, b).By symmetry, we may suppose that f (c) = M. (That is, if we can prove the theorem when f (c) = M, then a similar proof establishes the theorem when f (c) = m.)

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### Proof:

By the Extreme Value Theorem, f has a finite maximum M and a finite minimum m on [a,b]. If M = m, then f is

constant on (a,b) and f’(x)=0 for all x ∈ (a, b). Suppose that M 6= m. Since f (a) = f (b),f must assume one of the values M or m at some point c ∈ (a, b). By symmetry,we may suppose that f (c) = M. (That is, if we can prove the theorem when f (c) = M, then a similar proof establishes the theorem when f (c) = m.)

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### Proof:

By the Extreme Value Theorem, f has a finite maximum M and a finite minimum m on [a,b]. If M = m, then f is

constant on (a,b) and f’(x)=0 for all x ∈ (a, b). Suppose that M 6= m. Since f (a) = f (b), f must assume one of the values M or m at some point c ∈ (a, b).By symmetry, we may suppose that f (c) = M.(That is, if we can prove the theorem when f (c) = M, then a similar proof establishes the theorem when f (c) = m.)

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### Proof:

By the Extreme Value Theorem, f has a finite maximum M and a finite minimum m on [a,b]. If M = m, then f is

constant on (a,b) and f’(x)=0 for all x ∈ (a, b). Suppose that M 6= m. Since f (a) = f (b), f must assume one of the values M or m at some point c ∈ (a, b). By symmetry,we may suppose that f (c) = M. (That is, if we can prove the theorem when f (c) = M,then a similar proof establishes the theorem when f (c) = m.)

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### Proof:

By the Extreme Value Theorem, f has a finite maximum M and a finite minimum m on [a,b]. If M = m, then f is

constant on (a,b) and f’(x)=0 for all x ∈ (a, b). Suppose that M 6= m. Since f (a) = f (b), f must assume one of the values M or m at some point c ∈ (a, b). By symmetry, we may suppose that f (c) = M.(That is, if we can prove the theorem when f (c) = M, then a similar proof establishes the theorem when f (c) = m.)

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### Proof:

By the Extreme Value Theorem, f has a finite maximum M and a finite minimum m on [a,b]. If M = m, then f is

constant on (a,b) and f’(x)=0 for all x ∈ (a, b). Suppose that M 6= m. Since f (a) = f (b), f must assume one of the values M or m at some point c ∈ (a, b). By symmetry, we may suppose that f (c) = M. (That is, if we can prove the theorem when f (c) = M,then a similar proof establishes the theorem when f (c) = m.)

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### Proof:

By the Extreme Value Theorem, f has a finite maximum M and a finite minimum m on [a,b]. If M = m, then f is

constant on (a,b) and f’(x)=0 for all x ∈ (a, b). Suppose that M 6= m. Since f (a) = f (b), f must assume one of the values M or m at some point c ∈ (a, b). By symmetry, we may suppose that f (c) = M. (That is, if we can prove the theorem when f (c) = M, then a similar proof establishes the theorem when f (c) = m.)

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Since M is the maximum of f on [a, b], we have f (c + h) − f (c) ≤ 0

for all h that satisfy c + h ∈ (a, b).In this case h > 0 this implies that

f0(c) = lim

h→0+

f (c + h) − f (c)

h ≤ 0,

and in the case h < 0 this implies that f0(c) = lim

h→0−

f (c + h) − f (c)

h ≥ 0.

It follows that f0(c) = 0.2

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Since M is the maximum of f on [a, b],we have f (c + h) − f (c) ≤ 0

for all h that satisfy c + h ∈ (a, b). In this case h > 0 this implies that

f0(c) = lim

h→0+

f (c + h) − f (c)

h ≤ 0,

and in the case h < 0 this implies that f0(c) = lim

h→0−

f (c + h) − f (c)

h ≥ 0.

It follows that f0(c) = 0.2

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Since M is the maximum of f on [a, b], we have f (c + h) − f (c) ≤ 0

for all h that satisfy c + h ∈ (a, b).In this case h > 0 this implies that

f0(c) = lim

h→0+

f (c + h) − f (c)

h ≤ 0,

and in the case h < 0 this implies that f0(c) = lim

h→0−

f (c + h) − f (c)

h ≥ 0.

It follows that f0(c) = 0.2

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Since M is the maximum of f on [a, b], we have f (c + h) − f (c) ≤ 0

for all h that satisfy c + h ∈ (a, b). In this case h > 0 this implies that

f0(c) = lim

h→0+

f (c + h) − f (c)

h ≤ 0,

and in the case h < 0 this implies that f0(c) = lim

h→0−

f (c + h) − f (c)

h ≥ 0.

It follows that f0(c) = 0.2

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Since M is the maximum of f on [a, b], we have f (c + h) − f (c) ≤ 0

for all h that satisfy c + h ∈ (a, b). In this case h > 0 this implies that

f0(c) = lim

h→0+

f (c + h) − f (c)

h ≤ 0,

and in the case h < 0 this implies that f0(c) = lim

h→0−

f (c + h) − f (c)

h ≥ 0.

It follows that f0(c) = 0.2

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Since M is the maximum of f on [a, b], we have f (c + h) − f (c) ≤ 0

for all h that satisfy c + h ∈ (a, b). In this case h > 0 this implies that

f0(c) = lim

h→0+

f (c + h) − f (c)

h ≤ 0,

and in the case h < 0 this implies that f0(c) = lim

h→0−

f (c + h) − f (c)

h ≥ 0.

It follows that f0(c) = 0.2

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Theorem

Suppose a, b ∈R with a 6= b.

(i)[Generalized Mean Value Theorem]

If f,g are continuous on [a,b] and differentiable on (a,b), then there is a c ∈ (a, b) such that

g0(c)(f (b) − f (a)) = f0(c)(g(b) − g(a)).

(ii)[Mean Value Theorem]

If f is continuous on [a,b] and differentiable on (a,b), then there is a c ∈ (a, b) such that

f (b) − f (a) = f0(c)(b − a)

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Theorem

Suppose a, b ∈R with a 6= b.

(i)[Generalized Mean Value Theorem]

If f,g are continuous on [a,b] and differentiable on (a,b), then there is a c ∈ (a, b) such that

g0(c)(f (b) − f (a)) = f0(c)(g(b) − g(a)).

(ii)[Mean Value Theorem]

If f is continuous on [a,b] and differentiable on (a,b), then there is a c ∈ (a, b) such that

f (b) − f (a) = f0(c)(b − a)

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Theorem

Suppose a, b ∈R with a 6= b.

(i)[Generalized Mean Value Theorem]

If f,g are continuous on [a,b] and differentiable on (a,b), then there is a c ∈ (a, b) such that

g0(c)(f (b) − f (a)) = f0(c)(g(b) − g(a)).

(ii)[Mean Value Theorem]

If f is continuous on [a,b] and differentiable on (a,b), then there is a c ∈ (a, b) such that

f (b) − f (a) = f0(c)(b − a)

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Theorem

Suppose a, b ∈R with a 6= b.

(i)[Generalized Mean Value Theorem]

If f,g are continuous on [a,b] and differentiable on (a,b), then there is a c ∈ (a, b) such that

g0(c)(f (b) − f (a)) = f0(c)(g(b) − g(a)).

(ii)[Mean Value Theorem]

If f is continuous on [a,b] and differentiable on (a,b), then there is a c ∈ (a, b) such that

f (b) − f (a) = f0(c)(b − a)

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Theorem

Suppose a, b ∈R with a 6= b.

(i)[Generalized Mean Value Theorem]

If f,g are continuous on [a,b] and differentiable on (a,b), then there is a c ∈ (a, b) such that

g0(c)(f (b) − f (a)) = f0(c)(g(b) − g(a)).

(ii)[Mean Value Theorem]

If f is continuous on [a,b] and differentiable on (a,b), then there is a c ∈ (a, b) such that

f (b) − f (a) = f0(c)(b − a)

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Theorem

Suppose a, b ∈R with a 6= b.

(i)[Generalized Mean Value Theorem]

If f,g are continuous on [a,b] and differentiable on (a,b), then there is a c ∈ (a, b) such that

g0(c)(f (b) − f (a)) = f0(c)(g(b) − g(a)).

(ii)[Mean Value Theorem]

If f is continuous on [a,b] and differentiable on (a,b), then there is a c ∈ (a, b) such that

f (b) − f (a) = f0(c)(b − a)

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### Proof:

(i)

Set h(x ) = f (x )(g(b) − g(a)) − g(x )(f (b) − f (a)).Since h0(x ) = f0(x )(g(b) − g(a)) − g0(x )(f (b) − f (a)), it is clear that h is continuous on [a,b], differentiable on (a,b), and h(a) = h(b). Thus, by Rolle’s Theorem, h0(c) = 0 for some c ∈ (a, b).

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### Proof:

(i)

Set h(x ) = f (x )(g(b) − g(a)) − g(x )(f (b) − f (a)). Since h0(x ) = f0(x )(g(b) − g(a)) − g0(x )(f (b) − f (a)),it is clear that h is continuous on [a,b], differentiable on (a,b), and h(a) = h(b). Thus, by Rolle’s Theorem, h0(c) = 0 for some c ∈ (a, b).

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### Proof:

(i)

Set h(x ) = f (x )(g(b) − g(a)) − g(x )(f (b) − f (a)).Since h0(x ) = f0(x )(g(b) − g(a)) − g0(x )(f (b) − f (a)), it is clear that h is continuous on [a,b], differentiable on (a,b), and h(a) = h(b). Thus, by Rolle’s Theorem, h0(c) = 0 for some c ∈ (a, b).

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### Proof:

(i)

Set h(x ) = f (x )(g(b) − g(a)) − g(x )(f (b) − f (a)). Since h0(x ) = f0(x )(g(b) − g(a)) − g0(x )(f (b) − f (a)),it is clear that h is continuous on [a,b], differentiable on (a,b), and h(a) = h(b). Thus,by Rolle’s Theorem, h0(c) = 0 for some c ∈ (a, b).

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### Proof:

(i)

Set h(x ) = f (x )(g(b) − g(a)) − g(x )(f (b) − f (a)). Since h0(x ) = f0(x )(g(b) − g(a)) − g0(x )(f (b) − f (a)), it is clear that h is continuous on [a,b], differentiable on (a,b), and h(a) = h(b). Thus, by Rolle’s Theorem,h0(c) = 0 for some c ∈ (a, b).

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### Proof:

(i)

Set h(x ) = f (x )(g(b) − g(a)) − g(x )(f (b) − f (a)). Since h0(x ) = f0(x )(g(b) − g(a)) − g0(x )(f (b) − f (a)), it is clear that h is continuous on [a,b], differentiable on (a,b), and h(a) = h(b). Thus,by Rolle’s Theorem, h0(c) = 0 for some c ∈ (a, b).

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### Proof:

(i)

Set h(x ) = f (x )(g(b) − g(a)) − g(x )(f (b) − f (a)). Since h0(x ) = f0(x )(g(b) − g(a)) − g0(x )(f (b) − f (a)), it is clear that h is continuous on [a,b], differentiable on (a,b), and h(a) = h(b). Thus, by Rolle’s Theorem,h0(c) = 0 for some c ∈ (a, b).

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### Proof:

(i)

Set h(x ) = f (x )(g(b) − g(a)) − g(x )(f (b) − f (a)). Since h0(x ) = f0(x )(g(b) − g(a)) − g0(x )(f (b) − f (a)), it is clear that h is continuous on [a,b], differentiable on (a,b), and h(a) = h(b). Thus, by Rolle’s Theorem, h0(c) = 0 for some c ∈ (a, b).

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(ii)

Set g(x ) = x and apply part (i).(For a geometric

interpretation of this result, see the opening paragraph of this section and Figure 4.3 above.) 2

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(ii)

Set g(x ) = x and apply part (i). (For a geometric

interpretation of this result,see the opening paragraph of this section and Figure 4.3 above.) 2

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(ii)

Set g(x ) = x and apply part (i).(For a geometric

interpretation of this result, see the opening paragraph of this section and Figure 4.3 above.) 2

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(ii)

Set g(x ) = x and apply part (i). (For a geometric

interpretation of this result,see the opening paragraph of this section and Figure 4.3 above.) 2

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(ii)

Set g(x ) = x and apply part (i). (For a geometric

interpretation of this result, see the opening paragraph of this section and Figure 4.3 above.) 2

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Theorem (Bernoulli’s Inequality)

Let α be a positive real number and δ ≥ −1. If 0 < α ≤ 1, then

(1 + δ)α ≤ 1 + αδ, and if α ≥ 1, then

(1 + δ)α ≥ 1 + αδ,

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Theorem (Bernoulli’s Inequality)

Let α be a positive real number and δ ≥ −1. If 0 < α ≤ 1, then

(1 + δ)α ≤ 1 + αδ, and if α ≥ 1, then

(1 + δ)α ≥ 1 + αδ,

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1.Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ. If δ > 0, then c > 1.

Since 0 < α ≤ 1, it follows that cα−1≤ 1(see Exercise 5, p. 134); hence, δcα−1≤ δ. On the other hand, if

−1 ≤ δ ≤ 0, then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1. Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ. If δ > 0, then c > 1.

Since 0 < α ≤ 1, it follows that cα−1≤ 1(see Exercise 5, p. 134); hence, δcα−1≤ δ. On the other hand, if

−1 ≤ δ ≤ 0, then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1.Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ. If δ > 0, then c > 1.

Since 0 < α ≤ 1, it follows that cα−1≤ 1(see Exercise 5, p. 134); hence, δcα−1≤ δ. On the other hand, if

−1 ≤ δ ≤ 0, then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1. Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ.If δ > 0, then c > 1.

Since 0 < α ≤ 1, it follows that cα−1≤ 1(see Exercise 5, p. 134); hence, δcα−1≤ δ. On the other hand, if

−1 ≤ δ ≤ 0, then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1. Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ. If δ > 0, then c > 1.

Since 0 < α ≤ 1, it follows that cα−1≤ 1(see Exercise 5, p. 134); hence, δcα−1≤ δ. On the other hand, if

−1 ≤ δ ≤ 0, then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1. Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ.If δ > 0, then c > 1.

Since 0 < α ≤ 1,it follows that cα−1≤ 1(see Exercise 5, p. 134); hence, δcα−1≤ δ. On the other hand, if

−1 ≤ δ ≤ 0, then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1. Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ. If δ > 0, then c > 1.

Since 0 < α ≤ 1, it follows that cα−1≤ 1(see Exercise 5, p. 134);hence, δcα−1≤ δ. On the other hand, if

−1 ≤ δ ≤ 0, then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1. Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ. If δ > 0, then c > 1.

Since 0 < α ≤ 1,it follows that cα−1≤ 1(see Exercise 5, p. 134); hence,δcα−1≤ δ. On the other hand, if

−1 ≤ δ ≤ 0, then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1. Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ. If δ > 0, then c > 1.

Since 0 < α ≤ 1, it follows that cα−1≤ 1(see Exercise 5, p. 134);hence, δcα−1≤ δ. On the other hand, if

−1 ≤ δ ≤ 0, then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1. Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ. If δ > 0, then c > 1.

Since 0 < α ≤ 1, it follows that cα−1≤ 1(see Exercise 5, p. 134); hence,δcα−1≤ δ. On the other hand,if

−1 ≤ δ ≤ 0, then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1. Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ. If δ > 0, then c > 1.

Since 0 < α ≤ 1, it follows that cα−1≤ 1(see Exercise 5, p. 134); hence, δcα−1≤ δ. On the other hand, if

−1 ≤ δ ≤ 0,then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1. Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ. If δ > 0, then c > 1.

Since 0 < α ≤ 1, it follows that cα−1≤ 1(see Exercise 5, p. 134); hence, δcα−1≤ δ. On the other hand,if

−1 ≤ δ ≤ 0, then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1. Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ. If δ > 0, then c > 1.

Since 0 < α ≤ 1, it follows that cα−1≤ 1(see Exercise 5, p. 134); hence, δcα−1≤ δ. On the other hand, if

−1 ≤ δ ≤ 0,then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1. Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ. If δ > 0, then c > 1.

Since 0 < α ≤ 1, it follows that cα−1≤ 1(see Exercise 5, p. 134); hence, δcα−1≤ δ. On the other hand, if

−1 ≤ δ ≤ 0, then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1. Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ. If δ > 0, then c > 1.

Since 0 < α ≤ 1, it follows that cα−1≤ 1(see Exercise 5, p. 134); hence, δcα−1≤ δ. On the other hand, if

−1 ≤ δ ≤ 0, then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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### Proof:

The proof of these inequalities are similar. We present the details only for the case 0 < α ≤ 1. Let f (x ) = xα.By the Mean Value Theorem,

f (1 + δ) = f (1) + αδcα−1

for some c between 1 and 1 + δ. If δ > 0, then c > 1.

Since 0 < α ≤ 1, it follows that cα−1≤ 1(see Exercise 5, p. 134); hence, δcα−1≤ δ. On the other hand, if

−1 ≤ δ ≤ 0, then cα−1 ≥ 1 and again δcα−1≤ δ.

Therefore,

(1 + δ)α =f (1 + δ) = f (1) + αδcα−1≤ 1 + αδ. 2

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Theorem (L’Hopital’s Rule)

Let a be an extended real number and I be an open interval that either contains a or has a as an endpoint.

Suppose that f and g are differentiable on I \ {a}, and g(x ) 6= 0 6= g0(x ) for all x ∈ I \ {a}. Suppose futher that

A := lim

x →a x ∈I

f (x ) = lim

x →a x ∈I

g(x )

is either 0 or ∞.If

B := lim

x →ax ∈I

f0(x ) g0(x ) exists as an extended real number, then

x →alim

x ∈I

f (x )

g(x ) = lim

x →ax ∈I

f0(x ) g0(x ).

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Theorem (L’Hopital’s Rule)

Let a be an extended real number and I be an open interval that either contains a or has a as an endpoint.

Suppose that f and g are differentiable on I \ {a}, and g(x ) 6= 0 6= g0(x ) for all x ∈ I \ {a}. Suppose futher that

A := lim

x →a x ∈I

f (x ) = lim

x →a x ∈I

g(x )

is either 0 or ∞. If

B := lim

x →ax ∈I

f0(x ) g0(x ) exists as an extended real number, then

lim f (x )

= lim f0(x ) .

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Theorem (L’Hopital’s Rule)

Let a be an extended real number and I be an open interval that either contains a or has a as an endpoint.

Suppose that f and g are differentiable on I \ {a}, and g(x ) 6= 0 6= g0(x ) for all x ∈ I \ {a}. Suppose futher that

A := lim

x →a x ∈I

f (x ) = lim

x →a x ∈I

g(x )

is either 0 or ∞.If

B := lim

x →ax ∈I

f0(x ) g0(x ) exists as an extended real number, then

x →alim

x ∈I

f (x )

g(x ) = lim

x →ax ∈I

f0(x ) g0(x ).

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Theorem (L’Hopital’s Rule)

Let a be an extended real number and I be an open interval that either contains a or has a as an endpoint.

Suppose that f and g are differentiable on I \ {a}, and g(x ) 6= 0 6= g0(x ) for all x ∈ I \ {a}. Suppose futher that

A := lim

x →a x ∈I

f (x ) = lim

x →a x ∈I

g(x )

is either 0 or ∞. If

B := lim

x →ax ∈I

f0(x ) g0(x ) exists as an extended real number, then

lim f (x )

= lim f0(x ) .

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### Example:

Prove that the sequence (1 + 1

n)n is increasing, as n → ∞, and its limit e satisfies 2 < e ≤ 3 and log e = 1.

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### Example:

Prove that the sequence (1 + 1

n)n is increasing, as n → ∞, and its limit e satisfies 2 < e ≤ 3 and log e = 1.

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Find L = lim

x →1(log x )1−x

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Find L = lim

x →1(log x )1−x

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### Exercise:

Evaluate the following limits.

(a) lim

x →0+xx (b) lim

x →1

log x sin(πx )

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### Exercise:

Evaluate the following limits.

(a) lim

x →0+xx (b) lim

x →1

log x sin(πx )

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### Exercise:

Evaluate the following limits.

(a) lim

x →0+xx (b) lim

x →1

log x sin(πx )

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### Exercise:

Evaluate the following limits.

(a) lim

x →0+xx (b) lim

x →1

log x sin(πx )

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