Libro Calculo Diferencial Germán Rojas PDF

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libro de calculo diferencial que abarca con formalidad todos sus apuntes...

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ESCUELA POLITÉCNICA NACIONAL

Impreso en: Editogran-Medios Públicos EP.

E2

15 7

2.14

π

√ 2

4 3

4 3

4 3

1.31; 1.32; 1.33; 1.34; 1.35. 4 3

4 3 4 3

1.31

4 4 131 − 1.31 = − 100 3 3 400 393 7 = = − . 300 300 300 1.32

4 3

1.31 4 3

7 300

132 4 4 − 1.32 = − 3 100 3 4 396 400 = − . = 300 300 300 4 3

1.32 4 3

1.31

4 3

4 300

1.32

4 3

1.31 7 300

1.33

133 4 4 − 1.33 = − 3 100 3 1 399 400 = − . = 300 300 300 4 3

1.33

1 300

4 3

4 3

1.34

1.31

1.32

134 4 4 − 1.34 = − 3 100 3 400 2 402 = =− − . 300 300 300

4 3

       4  − 1.34  = − 2  = 2 .   300 300 3

4 3

1.34

4 3

1.35

    4     − 1.35  =  4 − 135   3 3 100    400 405  = 5 . − =  300 300 300 1.35

1.31

4 3

1.33 4 3

x

|x − y|

y

x y x y

x x

y

y

t P

O

t

P P P

y = x2

y = 2x − 1

y

y y = x3

(1, 1) y = 0.75x − 0.25x x

x

x=1

y = x2 (1, 1)

y = x3

x=1

y = 0.75x − 0.25

(0.5, 1.25) (0.5, 1.25)

C t

C

P P C Q Q1 Q2 Q3

t P

t

P

C

P

Q

P

Q1 Q2

P P t P

Q3

C y = g(x), g (a, g (a))

P C

t

P

P

t Q

P

t

R P R

Q

C R P R P

x

R

(x, g (x)) y

g(x)

R

P

g(a)

S

a

x

x

P

R ∠SP R

RS PS P

△SP R mx mx =

g(x) − g(a) R ∠SP R

x − a 6= 0

g (x) − g (a) . x−a R

P

(x, g (x))

P R

R

a C

P

x

P P P

R

P

g (x) − g (a) x−a x

a

y = 3x2

C (2, 12)

g(x) = 3x mx =

P

2

3x2 − 12 g(x) − g (2) = x−2 x−2

x

x 6= 2

2

(2, 12)

x 6= 2 mx =

3x2 − 12 3(x − 2)(x + 2) . = x−2 x−2 mx = 3(x + 2),

x 6= 2

x

mx mx

x x x R

mx

2 P x R

P x

2 mx R

P

P R

x 6= 2 mx = 3(x + 2). x=2 3(x + 2) = 3(2 + 2) = 12;

mx

x

2

12

mx

x

mx x

2

12 y = 3x2

(2, 12) 12

m

(a, b)

y − b = m(x − a). m = 12 y = 3x2

(a, b) = (2, 12) (2, 12) y − 12 = 12(x − 2),

y = 12x − 12.

y = 3x2 mx = x

2

(2, 12)

3x2 − 12 = 3(x + 2), x−2

12

y = 12x − 12.

y = 12x − 12 (2, 12)?

y = 3x2

y = x3

C s (1, 1)

1 C (s, s3 )

ms C

(1, 1)

ms s 1

ms

C

s ∈ {1.1, 1.5, 2} f (x) = xi = a ± 10−i f (x)

f (xi )

i ∈ {1, 2, . . . , 5} x

1 , (x − 2)2

f (x) = sin

a f (x)

a = 0.

a x

2

3 + 2x − x x2 − 4x + 7

x < 1, x > 1,

a=1

 2  x − 4x + 5

x < 1, x = 1, x > 1,

a = 1.

f (x) =

,

a

®

3  x+3

f (x) = √

3x − 15 , x2 − 10x + 25

f (x) =

sin(3x) , 2x

h h(x) =

x

√ 3

x3 + 8 − 2 x3

a = 5.

h(x) 0

x

a = 0.

mx = 12

x

x a

f (x) =

Äπ ä

a = 2.

3x2 − 12 x−2

2 x 6= 2

x=2 mx = 3(x + 2). mx = 12

x 6= 2

3(x + 2) = 12. x = 2. x 6= 2

x 6= 2 mx 6= 12. x

12

2 mx

12 mx mx 10−2

x 12 0.01

x

2 mx 12

10−2 mx

12

mx

10−2 |mx − 12| < 10−2 . x 2

x x

x

2

2 |x − 2|.

x

x x

2

2 x

2 |x − 2|

2 0 < |x − 2|. x 6= 2

x

2 |mx − 12| < 10−2

x

2

|mx − 12| < 10−2

|x − 2| |mx − 12| < 10−2

δ>0 0 < |x − 2| < δ,

|mx − 12| < 10−2

x 6= 2 mx = 3(x + 2), δ>0 0 < |x − 2| < δ

|mx − 12| < 10−2

12

δ δ x

x 6= 2 |x − 2| < δ,

|mx − 12| = |3(x + 2) − 12| < 10−2 . δ

δ

|3(x + 2) − 12| = |3x + 6 − 12| = |3x − 6|

= 3|x − 2|.

|3(x + 2) − 12| = 3|x − 2|. |x − 2| < δ. 3|x − 2| < 3δ.

|3(x + 2) − 12| < 3δ.

δ>0

x 6= 2

|x − 2| < δ,

|3(x + 2) − 12| < 3δ.

0 < |x − 2| < δ,

|3(x + 2) − 12| < 3δ. δ

|3(x + 2) − 12| < 10−2 .

10−2

3δ

|3(x + 2) − 12| < 3δ ≤ 10−2 ; 3δ ≤ 10−2 ,

|3(x + 2) − 12| < 10−2 , 3δ ≤ 10−2 δ≤

10−2 , 3

δ≤

10−2 3

δ>0

x 6= 2 |x − 2| < δ, |3(x + 2) − 12| < 10−2 .

00

|3(x + 2) − 12| < 10−2 .

|x − 2| < δ, x 6= 2

x 6= 2

|3(x + 2) − 12|

|x − 2| < δ

|3(x + 2) − 12| < 3δ. δ>0

3δ ≤ 10

−2

,

δ δ≤

10−2 . 3

δ x 6= 2

|x − 2| < δ |3(x + 2) − 12| < 10−2 .

2

x

mx

10−2

12

10−2

2

x

3 x 10−2

2

mx

mx

10−2

10−6

12

mx 12 10−6

10−2

x 12

12 10−6

10−2

12

10−6 mx 3

2

x

10−2

12

mx

2

mx

ǫ

ǫ ǫ

2

x

12

mx

3

ǫ

ǫ x |x − 2| <

ǫ , 3

|mx − 12| < ǫ. x

mx

12 mx

12

x mx 12

2 ǫ

x

12

mx x 2

2

2

ǫ 3

12

mx

2

x 12 = l´ım mx . x→2

mx

12

x

2 mx x

2

2

12

mx

ǫ>0 ǫ 3

δ>0

12

ǫ δ

mx x

2

2

ǫ>0

δ>0 |mx − 12| < ǫ, 0 < |x − 2| < δ.

12 = l´ım mx . x→2

a

L

I

a

f

I a

L

2 L f (x)

12 I

I ⊂ Dm(f ) ∪ {a}

a (−∞, +∞) x

f (x)

mx a

f (x)

L x a

a

f (x)

L

x

a

L = l´ım f (x), x→a

ǫ>0 f (x) x

δ>0

L ǫ

a

a

δ

Definición 1.1 (Límite de una función) Sean: 1. a y L dos números reales, 2. I un intervalo abierto que contiene el número a, y 3. f una función real definida en I , salvo, tal vez, en a; es decir, I ⊂ Dm(f ) ∪ {a}. Entonces:

L = l´ım f (x) x→a

si y solo si para todo ǫ > 0, existe un δ > 0 tal que

|f (x) − L| < ǫ, siempre que

0 < |x − a| < δ.

9 = l´ım x2 . x→−3

x2

9 x

−3 ǫ>0

δ>0 |x2 − 9| < ǫ x 6= −3

|x + 3| < δ.

δ

x2

9

|x2 − 9| = |(x − 3)(x + 3)| = |x − 3||x + 3|. x 6= −3

−3 |x − 3||x + 3| < ǫ, x 6= −3

−3

|x − 3||x + 3|

0 x

−3 0

|x − 3|

x

−3 −3

0

|x + 3|

|x − 3| x

−3 −4

x −3

−2

−1

−3

0

x −4 < x < −2, −1 < x + 3 < 1, |x + 3| < 1.

x−3

|x − 3|

|x − 3| −6

−7 < x − 3 < −5. x 6= −3 |x + 3| < 1, x−3 |x − 3| = −(x − 3); x − 3 = −|x − 3|, −7 < −|x − 3| < −5,

x 6= −3 7

7 > |x − 3| > 5. |x + 3| < 1

|x − 3| < 7.

|x − 3|

x 6= −3 |x+3| < 1 |x2 − 9| = |x − 3||x + 3| < 7|x + 3|, |x2 − 9| < 7|x + 3|, x 6= −3 x 6= −3

|x + 3| < 1.

|x + 3| < 1 7|x + 3| < ǫ,

|x2 − 9| < ǫ, x2

9

ǫ

x

x ǫ |x + 3| < . 7 |x2 − 9| < ǫ x 6= −3 |x + 3| <

|x + 3| < 1

ǫ . 7

δ > 0 δ ǫ 7

1

¶

δ = m´ın 1, δ≤1 x 6= −3

|x + 3| < δ

ǫ© . 7

δ≤

ǫ , 7

|x + 3| < δ, |x + 3| < δ ≤ 1

|x + 3| < δ ≤

ǫ , 7

|x2 − 9| < ǫ.

ǫ>0 |x2 − 9| < ǫ, x 6= −3

¶

|x + 3| < m´ın 1,

x2

9 x

ǫ© . 7

−3

2 9 = l´ım x . x→−3

L

f (x)

x δ

a ǫ

x2

9

|x2 − 9| = |x − 3||x + 3|. −3

x x |x + 3|

δ |x2 − 9| ǫ |x − 3| |x − 3| < 7, −3

|x + 3| < 1

x x 3

1

|x2 − 9| < 7|x + 3|, 0 < |x + 3| < δ |x2 − 9| < ǫ δ

|x + 3| < 1 0 < |x + 3| < δ n ǫo δ = m´ın 1, . 7

δ |x2 − 9| = |x − 3||x + 3| < 7|x + 3|, < 7δ, ǫ ≤ 7 = ǫ, 7

|x + 3| < δ ≤ 1,

|x + 3| < δ, ǫ δ≤ . 7

|x2 − 9| < ǫ, n ǫo 0 < |x + 3| < δ = m´ın 1, . 7

L = l´ım f (x). x→a

f (x) x 6= a |f (x) − L| = |g(x)||x − a|. |g(x)|

M >0

|g(x)| < M. M 0 < |x − a| < δ1 , δ1 > 0 a a

x

|f (x) − L| < M |x − a| < ǫ, 0 < |x − a| <

ǫ M

|x − a| < δ1 δ n

δ = m´ın δ1 ,

ǫ o . M

δ |f (x) − L| = |g(x)||x − a| < M |x − a|, |g(x)| < M < M δ, |x − a| < δ, ǫ ǫ . = ǫ, δ≤ ≤M M M |f (x) − L| < ǫ, n ǫo 0 < |x − a| < δ = m´ın δ1 , . M

2 = l´ım

x→3

5x − 3 . x+3

2 5x − 3 x+3 x 6= 3

3

|x − a| < δ ≤ δ1 ,

L

ǫ>0

δ>0

  5x − 3   − 2 < ǫ  x+3

x 6= 3

|x − 3| < δ. 5x−3 x+3

2

x 6= 2

  5x − 3    − 2 = |g(x)||x − 3|. x+3

     5x − 3   5x − 3 − 2(x + 3)  − 2 =    x+3 x+3      3x − 9  = 3  x − 3  =    x+3

x+3 |x − 3| . =3 |x + 3|

   5x − 3  − 2 =  x+3

g(x) =

3 |x − 3|. |x + 3| 3 , x+3 g(x)

M

δ1 > 0 3 < M, |x + 3| 0 < |x − 3| < δ1 . x

3 3 0

x

1

1

2

3

4

x 2 < x < 4, −1 < x − 3 < 1, |x − 3| < 1. M 3 5 < x + 3 < 7.

g(x)

x+3> 0 x + 3 = |x + 3|, 5 < |x + 3| < 7. |x + 3| > 0 1 , |x + 3| 1 1 1 > . > 5 |x + 3| 7 3 3 3 3 > . > 5 |x + 3| 7 |g(x)| =

|x − 3| < 1. 3 5

M x

3 3 < , 5 |x + 3|

δ1

1

|x − 3| < 1

  3 3   5x − 3 |x − 3| < |x − 3|,  − 2 = |x + 3| 5 x+3    5x − 3  3  − 2  < |x − 3| < ǫ, 5

x+3

0 < |x − 3| <

|x − 3| < 1

5 ǫ. 3

δ 5 δ = m´ın 1, ǫ . 3

n

o

δ 5x − 3 x+3 2

ǫ 3 |x − 3| |x + 3| 3 < |x − 3|, |x − 3| < δ ≤ 1, 5 3 < δ, |x − 3| < δ, 5 3 5 5 ≤ × ǫ = ǫ, δ ≤ ǫ. 3 5 3

   5x − 3   − 2 = x+3

   5x − 3  − 2 < ǫ  x+3

3 3 < , 5 |x + 3|

x 5 0 < |x − 3| < δ = m´ın 1, ǫ . 3

n

o

2 5x − 3 x+3 x

3

a>0 l´ım √ a a

x→a

x ǫ>0

√

x=

√

a. √

x

δ>0 √ √ | x − a| < ǫ,

|x − a| < δ

x>0 g(x)

x 6= a x > 0 √ √ | x − a| = |g(x)||x − a|.

|x − a|

√ √ | x − a|, √

x+

√

a.

0 √ √ √ √ √ √ | x + a| | x − a| = | x − a| × √ √ | x + a| √ √ |( x)2 − ( a)2 | = √ √ | x + a| |x − a| = √ √ . x+ a √ √ 1 √ |x − a|. | x − a| = √ x+ a g(x) = √ √

1 √ , x+ a √

g(x) x>0

√

x+

√

a>

√

a,

a > 0

x>0 √

1 1 √ 0 M

x>0

1 M=√ . a √ √ 1 | x − a| < √ |x − a|, a δ>0 |x − a| < δ, √ √ 1 δ | x − a| < √ |x − a| < √ . a a δ δ √ = ǫ. a √ δ = ǫ a. √ √ 1 √ |x − a| | x − a| = √ x+ a 1 < √ |x − a| a 1 < √ δ a √ ǫ a = √ = ǫ. a

x

√ √ | x − a| < ǫ √ 0 < |x − a| < δ = ǫ a. √ √ a x

x

a

|f (x) − L| ≤ M |x − a| M =

f g

√1 a

M

f

δ1

g

δ δ=

ǫ . M δ

ǫ

δ

ǫ

ǫ

δ 9 = l´ım x2 , x→−3

n ǫo δ = m´ın 1, . 7 ǫ = 14 ß ™ 14 δ = m´ın 1, = m´ın{1, 2} = 1. 7 ǫ=

7 2

ß δ = m´ın 1,

7 2×7

™

1 1 = m´ın{1, } = . 2 2 x2 x

9 14 −3

1

1 2

−3

x δ

δ(ǫ)

ǫ

δ

δ δ

ǫ

ǫ

f

R

x2

9 7 2

R

® f (x) =

x2 + 1 −2x2 + 8x − 4

x 1.

l´ım f (x) = 2

x→1

f

[−1.5, 3]

y 4 3 2 1

−1

1

2

f (x) 1 ǫ>0

x

3

2

x

δ>0 |f (x) − 2| < ǫ

x 6= 1

|x − 1| < δ

δ

ǫ

g

f (x)

x 6= 1 |f (x) − 2| = |g(x)||x − 1|. f 1

1 x1 x

1 1 00 C L

f (x)

x

a f (x) L+ǫ L L−ǫ

a−δ C

a

a+δ x f

(a, L)

(x, y ) (x, y ) f

(x, f (x)) (x, y )

(x, y ) f

f (x)

x C (a, L)

f 2ǫ

L f

C

2δ

f (x)

x

a

(a, L) f

(a, L)

f (x)

f (x)

L+ǫ L L−ǫ

L+ǫ L L−ǫ

a−δaa+δ

x

a−δ a a+δ f

x (a, L)

f g

f (x)

g(x)

h(x)

3

3

3

h

1 1

x

1

x

f (x) = x2

f : R −→ R δ>0

x∈R |f (x) − 9| < ǫ =

0 < |x − 3| < δ

9 = l´ım x2 . x→3

ǫ>0

¶

δ = m´ın 1,

ß δ = m´ın 1,

|x − 3| <

ǫ© 7

|x2 − 9| < ǫ

|x − 3| < δ ǫ = 21

1 14

|x2 − 9| <

1 2

7 1 2

1

™ =

1 . 14

x

1 2

9 = l´ımx→3 x2

1 2

ǫ=

C = (x, y) ∈ R2 : x ∈ 3 −

 1 1  1  1 , y ∈ 9− ,9+ ,3+ 2 2 14 14

f

(3, 9)





y 9+

1 2

9

9−

1 2

6

3

1

3−

2

1 14

3+ 3

9 = l´ım x2 , x→3

]3 −

f

1 14 , 3

9−

f (x)



f 3−

9−



f 3−

1 14



1 14







9+ 1 14



x = 3−

1 14



1 98 1 =8+ =8+ . 2 2 196 =8+

1 98 113 =9− . > 8+ 2 196 196 9+

f 3+



1 2

41 14 113 1 681 , = =8+ 196 196

=f

f (x)

f 3−

+ 141 [

1 14



1 2

x = 3+

43 14 85 1 849 , =9+ = 196 196

=f





1 1 98 . =9+ =9+ 196 2 2 =9+

1 98 85 =9+ . < 9+ 2 196 196

1 14

1 14

x

f

[0, +∞[ 9−

1 1 < f(x) < 9 + 2 2

3−

1 1 < x < 3+ . 14 14

f y 9+

y = x2

1 2

9

9−

1 2

6

3

1

1 3 − 14

2

L

3+ 3

f (x)

1 14

x

x

a L ǫ > 0

δ > 0

x 0 < |x − a| < δ

l´ım f (x) 6=

x→1

® f (x) =

x x+1

3 2

3 2 x ∈ [0, 1], x ∈ ]1, 2].

f (x)

® f (x) =

|f (x) − L| ≥ ǫ.

x x+1

x ∈ [0, 1], x ∈ ]1, 2],

x

1

f

ǫ>0

δ>0

x ∈ Dm(f )

  3  f (x) −  ≥ ǫ.

0 < |x − 1| < δ

2

f f (x)

2 3 2

1

1

x f

y=1 2ǫ

y=2

f 3 2

ǫ

1 4

ǫ

ǫ 3 2

3 2

f (x)

2 3 2

1

1

x 1

f f (x)

f (x)

2ǫ f (x)

2

2

2

3 2

3 2

3 2

1

1

1

1−δ

1

1+δ

x

1−δ 1

x

1+δ

1−δ 1 1+δ

x