• 沒有找到結果。

N/A
N/A
Protected

Copied!
54
0
0

(1)

WEN-CHING LIEN

Department of Mathematics National Cheng Kung University

(2)

## 3.1 Two-Sided Limits

Definition

Let aR, Let I be an open interval that contains a, and let f be a real function defined everywhere on I except possibly at a. Then f(x)is said to converge to L, as x approaches a, if and only if for everyǫ >0 there is aδ >0 (which in general depends on ǫ, f, I and a) such that

0< |xa| < δ implies |f(x) −L| < ǫ.

In this case we write

L= limf(x)

(3)

## 3.1 Two-Sided Limits

Definition

Let aR, Let I be an open interval that contains a, and let f be a real function defined everywhere on I except possibly at a. Then f(x)is said to converge to L, as x approaches a, if and only if for everyǫ >0 there is aδ >0 (which in general depends on ǫ, f, I and a) such that

0< |xa| < δ implies |f(x) −L| < ǫ.

In this case we write

L= lim

xaf(x)

and call L the limit of f(x)as x approaches a.

(4)

## 3.1 Two-Sided Limits

Definition

Let aR, Let I be an open interval that contains a, and let f be a real function defined everywhere on I except possibly at a. Then f(x)is said to converge to L, as x approaches a, if and only if for everyǫ >0 there is aδ >0 (which in general depends on ǫ, f, I and a) such that

0< |xa| < δ implies |f(x) −L| < ǫ.

In this case we write

L= limf(x)

(5)

### Example:

1. f(x) =3, lim

x→1f(x) =?

2. f(x) =3x, lim

x→1f(x) =?

3. f(x) =x2, lim

x→1f(x) =?

4. f(x) =√ x, lim

x→1f(x) =?

(6)

### Example:

1. f(x) =3, lim

x→1f(x) =?

2. f(x) =3x, lim

x→1f(x) =?

3. f(x) =x2, lim

x→1f(x) =?

4. f(x) =√ x, lim

x→1f(x) =?

(7)

### Example:

1. f(x) =3, lim

x→1f(x) =?

2. f(x) =3x, lim

x→1f(x) =?

3. f(x) =x2, lim

x→1f(x) =?

4. f(x) =√ x, lim

x→1f(x) =?

(8)

### Example:

1. f(x) =3, lim

x→1f(x) =?

2. f(x) =3x, lim

x→1f(x) =?

3. f(x) =x2, lim

x→1f(x) =?

4. f(x) =√ x, lim

x→1f(x) =?

(9)

### Example:

1. f(x) =3, lim

x→1f(x) =?

2. f(x) =3x, lim

x→1f(x) =?

3. f(x) =x2, lim

x→1f(x) =?

4. f(x) =√ x, lim

x→1f(x) =?

(10)

### Remark:

Let aR, let I be an open interval that contains a, and let f,g be real functions defined everywhere on I except

possibly at a. If f(x) = g(x)for all xI\ {a}and f(x) →L as xa, then g(x)also has a limit as xa, and

x→alimg(x) = lim

x→af(x).

(11)

### Remark:

Let aR, let I be an open interval that contains a, and let f,g be real functions defined everywhere on I except

possibly at a. If f(x) = g(x)for all xI\ {a}and f(x) →L as xa, then g(x)also has a limit as xa, and

x→alimg(x) = lim

x→af(x).

(12)

### Example:

g(x) = x3+x2x−1 x2−1 , lim

x1g(x) =?

(13)

### Example:

g(x) = x3+x2x−1 x2−1 , lim

x1g(x) =?

(14)

Theorem (Sequential Characterization of Limits)

Let aR, let I be an open interval that contains a, and let f be a real function defined everywhere on I except

possibly at a. Then

L= lim

x→af(x)

exists if and only if f(xn) →L as n→ ∞for every sequence xnI\ {a}that converges to a as n→ ∞.

(15)

Theorem (Sequential Characterization of Limits)

Let aR, let I be an open interval that contains a, and let f be a real function defined everywhere on I except

possibly at a. Then

L= lim

x→af(x)

exists if and only if f(xn) →L as n→ ∞for every sequence xnI\ {a}that converges to a as n→ ∞.

(16)

### Example:

Prove that

f(x) = (

sin(1

x) x 6=0

0 x =0

has no limit as x →0.

(17)

### Example:

Prove that

f(x) = (

sin(1

x) x 6=0

0 x =0

has no limit as x →0.

(18)

### Proof:

By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:

an:= 2

(4n+1)π and bn := 2

(4n+3)π, nN.

Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all nN, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2

(19)

### Proof:

By examing the graph of y =f(x)(see Figure 3.1),we are led to consider two extremes:

an:= 2

(4n+1)π and bn := 2

(4n+3)π, nN.

Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all nN, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2

(20)

### Proof:

By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:

an:= 2

(4n+1)π and bn := 2

(4n+3)π, nN.

Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all nN, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2

(21)

### Proof:

By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:

an:= 2

(4n+1)π and bn := 2

(4n+3)π, nN.

Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all nN, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2

(22)

### Proof:

By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:

an:= 2

(4n+1)π and bn := 2

(4n+3)π, nN.

Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all nN, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2

(23)

### Proof:

By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:

an:= 2

(4n+1)π and bn := 2

(4n+3)π, nN.

Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all nN, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2

(24)

### Proof:

By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:

an:= 2

(4n+1)π and bn := 2

(4n+3)π, nN.

Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all nN, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6,the limit of f(x), as x →0, cannot exist. 2

(25)

### Proof:

By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:

an:= 2

(4n+1)π and bn := 2

(4n+3)π, nN.

Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all nN, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2

(26)

### Proof:

By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:

an:= 2

(4n+1)π and bn := 2

(4n+3)π, nN.

Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all nN, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6,the limit of f(x), as x →0, cannot exist. 2

(27)

### Proof:

By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:

an:= 2

(4n+1)π and bn := 2

(4n+3)π, nN.

Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all nN, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2

(28)

Theorem

Suppose that aR, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at a. If f(x) and g(x) converge as x

approaches a, then so do (f +g)(x),(fg)(x),(αf)(x), and (f/g)(x)(when the limit of g(x) is nonzero).

In fact,

x→alim(f +g)(x) = lim

x→af(x) + lim

x→ag(x),

xlima(αf)(x) = αlim

xaf(x),

x→alim(fg)(x) = lim

x→af(x)lim

x→ag(x), and (when the limit of g(x) is nonzero)

(29)

Theorem

Suppose that aR, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at a. If f(x) and g(x) converge as x

approaches a, then so do (f +g)(x),(fg)(x),(αf)(x), and (f/g)(x)(when the limit of g(x) is nonzero).

In fact,

x→alim(f +g)(x) = lim

x→af(x) + lim

x→ag(x),

xlima(αf)(x) = αlim

xaf(x),

x→alim(fg)(x) = lim

x→af(x)lim

x→ag(x), and (when the limit of g(x) is nonzero)

x→alim

f g



(x) = limx→af(x) limxag(x).

(30)

Theorem

Suppose that aR, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at a. If f(x) and g(x) converge as x

approaches a, then so do (f +g)(x),(fg)(x),(αf)(x), and (f/g)(x)(when the limit of g(x) is nonzero).

In fact,

x→alim(f +g)(x) = lim

x→af(x) + lim

x→ag(x),

xlima(αf)(x) = αlim

xaf(x),

x→alim(fg)(x) = lim

x→af(x)lim

x→ag(x), and (when the limit of g(x) is nonzero)

(31)

Theorem

Suppose that aR, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at a. If f(x) and g(x) converge as x

approaches a, then so do (f +g)(x),(fg)(x),(αf)(x), and (f/g)(x)(when the limit of g(x) is nonzero).

In fact,

x→alim(f +g)(x) = lim

x→af(x) + lim

x→ag(x),

xlima(αf)(x) = αlim

xaf(x),

x→alim(fg)(x) = lim

x→af(x)lim

x→ag(x), and (when the limit of g(x) is nonzero)

x→alim

f g



(x) = limx→af(x) limxag(x).

(32)

Theorem

Suppose that aR, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at a. If f(x) and g(x) converge as x

approaches a, then so do (f +g)(x),(fg)(x),(αf)(x), and (f/g)(x)(when the limit of g(x) is nonzero).

In fact,

x→alim(f +g)(x) = lim

x→af(x) + lim

x→ag(x),

xlima(αf)(x) = αlim

xaf(x),

x→alim(fg)(x) = lim

x→af(x)lim

x→ag(x), and (when the limit of g(x) is nonzero)

(33)

Theorem (Squeeze Theorem For Functions) Suppose that aR, that I is an open interval that contains a, and that f,g,h are real functions defined everywhere on I except possibly at a.

(i)

If g(x) ≤h(x) ≤f(x)for all xI\ {a}, and

x→alimf(x) = lim

x→ag(x) =L, then the limit of h(x) exists, as xa, and

xlimah(x) =L.

(34)

Theorem (Squeeze Theorem For Functions) Suppose that aR, that I is an open interval that contains a, and that f,g,h are real functions defined everywhere on I except possibly at a.

(i)

If g(x) ≤h(x) ≤f(x)for all xI\ {a}, and

x→alimf(x) = lim

x→ag(x) =L, then the limit of h(x) exists, as xa, and

(35)

Theorem (Squeeze Theorem For Functions) Suppose that aR, that I is an open interval that contains a, and that f,g,h are real functions defined everywhere on I except possibly at a.

(i)

If g(x) ≤h(x) ≤f(x)for all xI\ {a}, and

x→alimf(x) = lim

x→ag(x) =L, then the limit of h(x) exists, as xa, and

xlimah(x) =L.

(36)

Theorem (Squeeze Theorem For Functions) Suppose that aR, that I is an open interval that contains a, and that f,g,h are real functions defined everywhere on I except possibly at a.

(i)

If g(x) ≤h(x) ≤f(x)for all xI\ {a}, and

x→alimf(x) = lim

x→ag(x) =L, then the limit of h(x) exists, as xa, and

(37)

Theorem (ii)

If|g(x)| ≤M for all xI\ {a}and f(x) →0 as xa, then

x→alimf(x)g(x) =0.

(38)

Theorem (ii)

If|g(x)| ≤M for all xI\ {a}and f(x) →0 as xa, then

x→alimf(x)g(x) =0.

(39)

Theorem (ii)

If|g(x)| ≤M for all xI\ {a}and f(x) →0 as xa, then

x→alimf(x)g(x) =0.

(40)

Theorem (Comparison Theorem For Functions) Suppose that aR, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at a. If f and g have a limit as x approaches a and

f(x) ≤g(x), xI\ {a}, then

x→alimf(x) ≤ lim

x→ag(x).

(41)

Theorem (Comparison Theorem For Functions) Suppose that aR, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at a. If f and g have a limit as x approaches a and

f(x) ≤g(x), xI\ {a}, then

x→alimf(x) ≤ lim

x→ag(x).

(42)

### Example:

For each function f define the positive part of f by f+(x) = |f(x)| +f(x)

2 , xDom(f), and the negative part by

f(x) = |f(x)| −f(x)

2 , xDom(f).

(43)

### Example:

For each function f define the positive part of f by f+(x) = |f(x)| +f(x)

2 , xDom(f), and the negative part by

f(x) = |f(x)| −f(x)

2 , xDom(f).

(44)

(a)

Prove that f+(x) ≥0, f(x) ≥0, f(x) =f+(x) −f(x), and

|f(x)| =f+(x) +f(x)hold for all xDom(f).(Compare with Exercise 1,p.11.)

(b)

Prove that if

L= lim

x→af(x)

exists, then f+(x) →L+and f(x) →L as xa.

(45)

(a)

Prove that f+(x) ≥0, f(x) ≥0, f(x) =f+(x) −f(x), and

|f(x)| =f+(x) +f(x)hold for all xDom(f).(Compare with Exercise 1,p.11.)

(b)

Prove that if

L= lim

x→af(x)

exists, then f+(x) →L+and f(x) →L as xa.

(46)

(a)

Prove that f+(x) ≥0, f(x) ≥0, f(x) =f+(x) −f(x), and

|f(x)| =f+(x) +f(x)hold for all xDom(f).(Compare with Exercise 1,p.11.)

(b)

Prove that if

L= lim

x→af(x)

exists, then f+(x) →L+and f(x) →L as xa.

(47)

(a)

Prove that f+(x) ≥0, f(x) ≥0, f(x) =f+(x) −f(x), and

|f(x)| =f+(x) +f(x)hold for all xDom(f).(Compare with Exercise 1,p.11.)

(b)

Prove that if

L= lim

x→af(x)

exists, then f+(x) →L+and f(x) →L as xa.

(48)

### Example:

Let f,g be real functions, and for each

xDom(f) ∩Dom(g)define(f∨g)(x) :=max{f(x),g(x)}

and(f ∨g)(x) :=min{f(x),g(x)}.

(49)

### Example:

Let f,g be real functions, and for each

xDom(f) ∩Dom(g)define(f∨g)(x) :=max{f(x),g(x)}

and(f ∨g)(x) :=min{f(x),g(x)}.

(50)

(a)

Prove that

(f ∨g)(x) = (f +g)(x) + |(fg)(x)|

2 and

(f ∧g)(x) = (f +g)(x) − |(fg)(x)|

2 for all xDom(f) ∩Dom(g).

(51)

(a)

Prove that

(f ∨g)(x) = (f +g)(x) + |(fg)(x)|

2 and

(f ∧g)(x) = (f +g)(x) − |(fg)(x)|

2 for all xDom(f) ∩Dom(g).

(52)

(b)

Prove that if

L= lim

xaf(x) and M = lim

xag(x)

exist, then (f ∨g)(x) →LM and(f ∧g)(x) →LM as xa.

(53)

(b)

Prove that if

L= lim

xaf(x) and M = lim

xag(x)

exist, then (f ∨g)(x) →LM and(f ∧g)(x) →LM as xa.

(54)

## Thank you.

In this section we define an integral that is similar to a single integral except that instead of integrating over an interval [a, b], we integrate over a curve C.. Such integrals

We summarize these properties as follows, using the fact that this function is just a special case of the exponential functions considered in Theorem 2 but with base b = e

One-to-one functions are important because they are precisely the functions that possess inverse functions according to the following definition.. This definition says that if f

— John Wanamaker I know that half my advertising is a waste of money, I just don’t know which half.. —

Suppose that E is bounded, open interval and that each f k.. is differentiable

— John Wanamaker I know that half my advertising is a waste of money, I just don’t know which half.. —

— John Wanamaker I know that half my advertising is a waste of money, I just don’t know which half.. —

• Suppose the input graph contains at least one tour of the cities with a total distance at most B. – Then there is a computation path for