Ejercicios Límites y Continuidad

Page 1

CALCULO DIFERENCIAL Marco Antonio HernĂĄndez MartĂ­nez MaestrĂ­a en EducaciĂłn MatemĂĄtica

Ejercicios de reforzamiento

I. Con base en las propiedades de límites calcular. 1) lim 5 �→0

đ??śđ?‘œđ?‘› đ?‘?đ?‘Žđ?‘ đ?‘’ đ?‘’đ?‘› đ?‘’đ?‘™ đ?‘Ąđ?‘’đ?‘œđ?‘&#x;đ?‘’đ?‘šđ?‘Ž lim đ?‘? = đ?‘?, đ?‘? = đ?‘?đ?‘Ąđ?‘’. đ?‘Ľâ†’đ?‘Ž

∴ lim 5 = 5 đ?‘Ľâ†’0

2) lim đ?‘Ľ đ?‘Ľâ†’−7

lim đ?‘Ľ = đ?‘Ž

�→�

∴ lim đ?‘Ľ = −7 đ?‘Ľâ†’−7

3) lim đ?‘Ľ 4 đ?‘Ľâ†’−7

lim đ?‘“(đ?‘Ľ) = đ??ż

�→�

lim đ?‘Ľ 4 = (−7)4 = 2041 đ?‘Ľâ†’−7

∴ lim đ?‘Ľ 4 = 2041 đ?‘Ľâ†’−7

4) lim 3đ?‘Ľ đ?‘Ľâ†’−4

lim 3đ?‘Ľ = 3 lim đ?‘Ľ = (3)(−4) = −12

đ?‘Ľâ†’−4

đ?‘Ľâ†’−4

∴ lim 3đ?‘Ľ = −12 đ?‘Ľâ†’−4

5) lim[3� + 2� 2 ] �→1

lim[3đ?‘Ľ + 2đ?‘Ľ 2 ] = lim 3đ?‘Ľ + lim 2đ?‘Ľ 2 = 3 lim đ?‘Ľ + 2 lim đ?‘Ľ 2 = (3)(1) + (2)(1)2 = 3 + 2 = 5

�→1

�→1

�→1

�→1

�→1

∴ lim[3đ?‘Ľ + 2đ?‘Ľ 2 ] = 5 đ?‘Ľâ†’1

6) lim[đ?‘Ľ(đ?‘Ľ 2 − 1)] đ?‘Ľâ†’2

lim[đ?‘Ľ(đ?‘Ľ 2 − 1)] = lim[đ?‘Ľ 3 − đ?‘Ľ] = limđ?‘Ľ 3 − limđ?‘Ľ = (2)3 − 2 = 8 − 2 = 6

�→2

�→2

�→2

�→2

∴ lim[đ?‘Ľ(đ?‘Ľ 2 − 1)] = 6 đ?‘Ľâ†’2

HernĂĄndez MartĂ­nez Marco Antonio

42800354

PĂĄgina 1

CALCULO DIFERENCIAL Marco Antonio HernĂĄndez MartĂ­nez MaestrĂ­a en EducaciĂłn MatemĂĄtica

đ?‘Ľâˆ’2 7) lim ( ) đ?‘Ľâ†’4 đ?‘Ľ lim (

�→4

lim (đ?‘Ľ − 2) lim đ?‘Ľ − lim 2 4 − 2 đ?‘Ľâˆ’2 2 1 đ?‘Ľâ†’4 ) = đ?‘Ľâ†’4 = đ?‘Ľâ†’4 = = = đ?‘Ľ lim đ?‘Ľ lim đ?‘Ľ 4 4 2 đ?‘Ľâ†’4

�→4

đ?‘Ľâˆ’2 1 ∴ lim ( )= đ?‘Ľâ†’4 đ?‘Ľ 2

8) lim ���(�) �→0.5

lim ���(�) = ��� (0.5) = 0.008

�→0.5

∴ lim đ?‘†đ?‘’đ?‘›(đ?‘Ľ) = 0.008 đ?‘Ľâ†’0.5

9) lim3 ���(�) �→

4

lim3 đ?‘†đ?‘’đ?‘› (đ?‘Ľ) đ?‘Ľâ†’ đ?‘†đ?‘’đ?‘› (đ?‘Ľ) 0.01308 4 lim3 = = = 0.0130 lim3 đ??śđ?‘œđ?‘ (đ?‘Ľ) 0.9999 đ?‘Ľâ†’ đ??śđ?‘œđ?‘ (đ?‘Ľ) 4

�→

4

∴ lim đ?‘‡đ?‘Žđ?‘›(đ?‘Ľ) = 0.0130 3 đ?‘Ľâ†’

4

2. Encuentra los siguientes lĂ­mites laterales. 1) lim+ √đ?‘Ľ − 3 đ?‘Ľâ†’3

HernĂĄndez MartĂ­nez Marco Antonio

đ?‘Ľ+

đ?‘“(đ?‘Ľ) = √đ?‘Ľ − 3

3.1

0.3162

3.01

0.1

3.001

0.03162

3.0001

0.01

3.00001

0.00316

3.000001

0.001

� →3

No tiene lĂ­mite

42800354

PĂĄgina 2

CALCULO DIFERENCIAL Marco Antonio Hernández Martínez Maestría en Educación Matemática

𝐿𝑖𝑚𝑖𝑡𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑝𝑜𝑟 𝑙𝑎 𝑖𝑧𝑞𝑢𝑖𝑒𝑟𝑑𝑎 𝑛𝑜𝑠 𝑞𝑢𝑒𝑑𝑎 𝑢𝑛𝑎 𝑟𝑎𝑖𝑧 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑎, 𝑝𝑜𝑟 𝑙𝑜 𝑡𝑎𝑛𝑡𝑜 𝑛𝑜 𝑡𝑖𝑒𝑛𝑒 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑒𝑛 𝑙𝑜𝑠 𝑟𝑒𝑎𝑙𝑒𝑠 → lim+ 𝑓(𝑥) ≠ lim− 𝑓(𝑥) 𝑥→3

𝑥→3

∴ lim √𝑥 − 3 𝑛𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑥→3

2) lim − (𝑥 + 6) 𝑥→−2

Hernández Martínez Marco Antonio

𝑥+

𝑓(𝑥) = 𝑥 + 6

-2.1

3.901

-2.01

3.99

-2.001

3.999

-2.0001

3.9999

-2.00001

3.99999

-2.000001

3.999999

↓

↓

-2

4

𝑥−

𝑓(𝑥) = 𝑥 + 6

-1.9

4.1

-1.99

4.01

-1.999

4.001

-1.9999

4.001

-1.99999

4.0001

42800354

Página 3

CALCULO DIFERENCIAL Marco Antonio HernĂĄndez MartĂ­nez MaestrĂ­a en EducaciĂłn MatemĂĄtica

-1.999999

4.00001

↓

↓

-2

4

→

lim (đ?‘Ľ + 6) =

đ?‘Ľâ†’−2+

lim (đ?‘Ľ + 6)

đ?‘Ľâ†’−2−

∴ lim (đ?‘Ľ + 6) = 4 đ?‘Ľâ†’−2

3) lim− đ?‘Ľ 3 đ?‘Ľâ†’0

đ?‘Ľâˆ’

đ?‘“(đ?‘Ľ) = đ?‘Ľ 3

-0.1

-0.01

-0.01

-0.000001

-0.001

-0.000000001

-0.0001

-0.000000000001

↓

↓

0

0

đ?‘Ľ+

đ?‘“(đ?‘Ľ) = đ?‘Ľ 3

0.1

0.01

0.01

0.000001

0.001

0.000000001

0.0001

0.000000000001

↓

↓

0

0

→ lim+ đ?‘Ľ 3 = lim− đ?‘Ľ 3 đ?‘Ľâ†’0

�→0

∴ lim đ?‘Ľ 3 = 0 đ?‘Ľâ†’0

HernĂĄndez MartĂ­nez Marco Antonio

42800354

PĂĄgina 4

CALCULO DIFERENCIAL Marco Antonio HernĂĄndez MartĂ­nez MaestrĂ­a en EducaciĂłn MatemĂĄtica

−đ?‘Ľ 2 | đ?‘Ľ < 0 4) lim+ { 2 đ?‘Ľâ†’2 đ?‘Ľ |đ?‘Ľ ≼0

đ?‘Ľ+

đ?‘“(đ?‘Ľ) = đ?‘Ľ 2

2.1

4.41

2.01

4.0401

2.001

4.004001

2.0001

4.00040001

↓

↓

2

4

đ?‘Ľâˆ’

đ?‘“(đ?‘Ľ) = đ?‘Ľ 2

1.9

3.61

1.99

3.9601

1.999

3.996001

1.9999

3.9996001

↓

↓

2

4

→ lim+ đ?‘“(đ?‘Ľ) = lim− đ?‘“(đ?‘Ľ) đ?‘Ľâ†’2

HernĂĄndez MartĂ­nez Marco Antonio

42800354

�→2

PĂĄgina 5

CALCULO DIFERENCIAL Marco Antonio Hernández Martínez Maestría en Educación Matemática

∴ lim 𝑥 2 = 4 𝑥→2

𝐶𝑜𝑚𝑜 𝑥 → 2, 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝑓(𝑥) = −𝑥 2 𝑛𝑢𝑛𝑐𝑎 𝑠𝑒 𝑣𝑎 𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑟, 𝑦𝑎 𝑞𝑢𝑒 𝑓(𝑥) = −𝑥 2 | 𝑥 < 0

5) lim−√𝑥 + 4 𝑥→4

𝑥−

𝑓(𝑥) = √𝑥 + 4

3.9

2.81

3.99

2.8266

3.999

2.8282

3.9999

2.828409

3.99999

2.828442

↓

↓

4

2.83

𝑥+

𝑓(𝑥) = √𝑥 + 4

4.1

2.8460

4.01

2.830194

4.001

2.828603

4.0001

2.828444

↓

↓

4

2.83

→ lim+ √𝑥 + 4 = lim− √𝑥 + 4 𝑥→4

Hernández Martínez Marco Antonio

42800354

𝑥→4

Página 6

CALCULO DIFERENCIAL Marco Antonio HernĂĄndez MartĂ­nez MaestrĂ­a en EducaciĂłn MatemĂĄtica

∴ lim √đ?‘Ľ + 4 ≈ 2.83 đ?‘Ľâ†’4

6) lim+ { �→0

đ?‘Ľ + 1| đ?‘Ľ < 0 −đ?‘Ľ + 5| đ?‘Ľ ≼ 0

HernĂĄndez MartĂ­nez Marco Antonio

đ?‘Ľ+

đ?‘“(đ?‘Ľ) = đ?‘Ľ + 1

0.1

1.1

0.01

1.01

0.001

1.001

0.0001

1.0001

0.00001

1.00001

↓

↓

0

1

đ?‘Ľâˆ’

đ?‘“(đ?‘Ľ) = −đ?‘Ľ + 5

-0.1

5.1

-0.01

5.01

-0.001

5.001

-0.0001

5.0001

-0.00001

5.00001

↓

↓

42800354

PĂĄgina 7

CALCULO DIFERENCIAL Marco Antonio HernĂĄndez MartĂ­nez MaestrĂ­a en EducaciĂłn MatemĂĄtica

0

5

→ lim+ đ?‘“(đ?‘Ľ) ≠lim− đ?‘“(đ?‘Ľ) đ?‘Ľâ†’0

�→0

∴ đ?‘’đ?‘™ đ?‘™đ?‘–đ?‘šđ?‘–đ?‘Ąđ?‘’ đ?‘›đ?‘œ đ?‘’đ?‘Ľđ?‘–đ?‘ đ?‘Ąđ?‘’

đ?‘Ľ+

đ?‘“(đ?‘Ľ) = −đ?‘Ľ + 5

0.1

4.9

0.01

4.99

0.001

4.999

0.0001

4.9999

0.00001

4.99999

↓

↓

0

5

đ?‘Ľâˆ’

đ?‘“(đ?‘Ľ) = đ?‘Ľ + 1

-0.1

0.9

-0.01

0.99

-0.001

0.999

-0.0001

0.9999

-0.00001

0.99999

↓

↓

0

1

→ lim+ đ?‘“(đ?‘Ľ) ≠lim− đ?‘“(đ?‘Ľ) đ?‘Ľâ†’0

�→0

∴ đ?‘’đ?‘™ đ?‘™đ?‘–đ?‘šđ?‘–đ?‘Ąđ?‘’ đ?‘‘đ?‘’ đ?‘“(đ?‘Ľ) đ?‘›đ?‘œ đ?‘’đ?‘Ľđ?‘–đ?‘ đ?‘Ąđ?‘’

HernĂĄndez MartĂ­nez Marco Antonio

42800354

PĂĄgina 8

CALCULO DIFERENCIAL Marco Antonio Hernández Martínez Maestría en Educación Matemática

3. Dadas las funciones encuentra los límites y asíntotas verticales. 1) lim 𝑥→?

𝑥+2 𝑥−3 𝑥−3 =0 → 𝑥 =3 𝐷𝑜𝑚 𝑓 = {𝑥0 ∈ ℝ | 𝑥0 ≠ 3 } 𝑡𝑖𝑒𝑛𝑒 𝑢𝑛𝑎 𝑎𝑠𝑖𝑛𝑡𝑖𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑒𝑛 𝑥 = 3

lim

𝑥→3

lim

𝑥→3

𝑥+2 = 𝑥−3

(𝑥 + 2) (𝑥 + 3) (𝑥 2 + 3𝑥 + 2𝑥 + 6) (𝑥 2 + 5𝑥 + 6) = lim 2 = lim 𝑥→3 (𝑥 + 3𝑥 − 3𝑥 − 9) 𝑥→3 (𝑥 − 3) (𝑥 + 3) (𝑥 2 − 9) 30 = 𝑁𝑜 𝑠𝑒 𝑝𝑢𝑒𝑑𝑒 𝑑𝑖𝑣𝑖𝑑𝑖𝑟 𝑒𝑛𝑡𝑟𝑒 0 0

𝑦 𝑎𝑑𝑒𝑚𝑎𝑠 𝑑𝑒 𝑙𝑎 𝑔𝑟𝑎𝑓𝑖𝑐𝑎 𝑠𝑒 𝑣𝑒 𝑞𝑢𝑒 𝑐𝑢𝑎𝑛𝑑𝑜 lim+ 𝑥→3

∴ lim

𝑥→3

2) lim = 𝑥→?

𝑥+2 𝑥+2 = +⋈; 𝑦 𝑐𝑢𝑎𝑛𝑑𝑜 lim− = −⋈ 𝑥→3 𝑥 − 3 𝑥−3

𝑥+2 =⋈ 𝑥−3

1 𝑥 𝐷𝑜𝑚 𝑓 = {𝑥0 ∈ ℝ | 𝑥0 ≠ 0 }

Hernández Martínez Marco Antonio

42800354

Página 9

CALCULO DIFERENCIAL Marco Antonio Hernández Martínez Maestría en Educación Matemática

𝐷𝑒 𝑙𝑎 𝑔𝑟𝑎𝑓𝑖𝑐𝑎 𝑠𝑒 𝑜𝑏𝑠𝑒𝑣𝑎 𝑞𝑢𝑒 𝑓(𝑥) =

1 𝑡𝑖𝑒𝑛𝑒 𝑢𝑛𝑎 𝑎𝑠𝑖𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑣𝑎𝑙 𝑒𝑛 𝑥 = 0 𝑥

𝑦 𝑎𝑑𝑒𝑚𝑎𝑠 𝑑𝑒 𝑙𝑎 𝑔𝑟𝑎𝑓𝑖𝑐𝑎 𝑑𝑒 𝑓 𝑠𝑒 𝑣𝑒 𝑞𝑢𝑒 𝑐𝑢𝑎𝑛𝑑𝑜 lim+ 𝑥→0

∴ lim 𝑥→0

4) lim 𝑥→?

1 1 = +⋈; 𝑦 𝑐𝑢𝑎𝑛𝑑𝑜 lim− = −⋈ 𝑥→0 𝑥 𝑥

1 =⋈ 𝑥

1 𝑥3 − 1 3

3 𝑥 3 − 1 = 0 → 𝑥 3 = 1 → √𝑥 3 = √1 → 𝑥 = 1

𝐷𝑜𝑚 𝑓 = {𝑥0 ∈ ℝ | 𝑥0 ≠ 1 }

Hernández Martínez Marco Antonio

42800354

Página 10

CALCULO DIFERENCIAL Marco Antonio Hernández Martínez Maestría en Educación Matemática

𝐷𝑒 𝑙𝑎 𝑔𝑟𝑎𝑓𝑖𝑐𝑎 𝑠𝑒 𝑜𝑏𝑠𝑒𝑣𝑎 𝑞𝑢𝑒 𝑓(𝑥) =

1 𝑡𝑖𝑒𝑛𝑒 𝑢𝑛𝑎 𝑎𝑠𝑖𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑣𝑎𝑙 𝑒𝑛 𝑥 = 1 𝑥3 − 1

𝑦 𝑎𝑑𝑒𝑚𝑎𝑠 𝑑𝑒 𝑙𝑎 𝑔𝑟𝑎𝑓𝑖𝑐𝑎 𝑑𝑒 𝑓 𝑠𝑒 𝑣𝑒 𝑞𝑢𝑒 𝑐𝑢𝑎𝑛𝑑𝑜 lim+ 𝑥→1

∴ lim 𝑥→1

5) lim 𝑥→?

𝑥3

1 1 = +⋈; 𝑦 𝑐𝑢𝑎𝑛𝑑𝑜 lim− 3 = −⋈ 𝑥→1 𝑥 − 1 −1

1 =⋈ 𝑥3 − 1

9 𝑥−9 𝑥−9 =0 → 𝑥 =9 𝐷𝑜𝑚 𝑓 = {𝑥0 ∈ ℝ | 𝑥0 ≠ 9 }

𝐷𝑒 𝑙𝑎 𝑔𝑟𝑎𝑓𝑖𝑐𝑎 𝑠𝑒 𝑜𝑏𝑠𝑒𝑣𝑎 𝑞𝑢𝑒 𝑓(𝑥) =

9 , 𝑡𝑖𝑒𝑛𝑒 𝑢𝑛𝑎 𝑎𝑠𝑖𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑣𝑎𝑙 𝑒𝑛 𝑥 = 9 𝑥−9

𝑦 𝑎𝑑𝑒𝑚𝑎𝑠 𝑑𝑒 𝑙𝑎 𝑔𝑟𝑎𝑓𝑖𝑐𝑎 𝑑𝑒 𝑓 𝑠𝑒 𝑣𝑒 𝑞𝑢𝑒 𝑐𝑢𝑎𝑛𝑑𝑜 lim+ 𝑥→9

∴ lim 𝑥→9

5) lim

𝑥→? 𝑥 2

9 9 = +⋈; 𝑦 𝑐𝑢𝑎𝑛𝑑𝑜 lim− = −⋈ 𝑥→9 𝑥 − 9 𝑥−9

9 =⋈ 𝑥−9

1 − 16

Hernández Martínez Marco Antonio

42800354

Página 11

CALCULO DIFERENCIAL Marco Antonio Hernández Martínez Maestría en Educación Matemática

𝑥 2 − 16 = 0 → (𝑥 − 4)(𝑥 + 4) = 0 → (𝑥 − 4) = 0 ∧ (𝑥 + 4) = 0 𝑥1 = 4 ∧ 𝑥2 = −4 𝐷𝑜𝑚 𝑓 = {𝑥0 ∈ ℝ | 𝑥0 ≠ −4, 4 }

𝐷𝑒 𝑙𝑎 𝑔𝑟𝑎𝑓𝑖𝑐𝑎 𝑠𝑒 𝑜𝑏𝑠𝑒𝑣𝑎 𝑞𝑢𝑒 𝑓(𝑥) =

𝑥2

1 , 𝑡𝑖𝑒𝑛𝑒 𝑑𝑜𝑠 𝑎𝑠𝑖𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑒𝑠 𝑢𝑛𝑎 𝑒𝑛 𝑥1 = −4, − 16

𝑥2 = 4 𝑦 𝑎𝑑𝑒𝑚𝑎𝑠 𝑑𝑒 𝑙𝑎 𝑔𝑟𝑎𝑓𝑖𝑐𝑎 𝑑𝑒 𝑓 𝑠𝑒 𝑣𝑒 𝑞𝑢𝑒 𝑐𝑢𝑎𝑛𝑑𝑜 lim+ 𝑥→4

lim

𝑥→9−

𝑥2

1 = +⋈ 𝑦 𝑐𝑢𝑎𝑛𝑑𝑜 − 16

1 1 1 = −⋈ 𝑦 lim + 2 = −⋈ 𝑦 𝑐𝑢𝑎𝑛𝑑𝑜 lim − 2 = +⋈ 𝑥→−4 𝑥 − 16 𝑥→−4 𝑥 − 16 𝑥 2 − 16 ∴ lim

𝑥→ −4 𝑥 2

6) lim 𝑥→?

1 1 =⋈ ∧ lim 2 =⋈ 𝑥→ 4 𝑥 − 16 − 16

3 𝑥−4 𝑥−4 =0 → 𝑥 =4 𝐷𝑜𝑚 𝑓 = {𝑥0 ∈ ℝ | 𝑥0 ≠ 4 }

Hernández Martínez Marco Antonio

42800354

Página 12

CALCULO DIFERENCIAL Marco Antonio Hernández Martínez Maestría en Educación Matemática

𝐷𝑒 𝑙𝑎 𝑔𝑟𝑎𝑓𝑖𝑐𝑎 𝑠𝑒 𝑜𝑏𝑠𝑒𝑣𝑎 𝑞𝑢𝑒 𝑓(𝑥) =

3 , 𝑡𝑖𝑒𝑛𝑒 𝑢𝑛𝑎 𝑎𝑠𝑖𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑒 𝑒𝑛 𝑥 = 4, 𝑥−4

𝑦 𝑎𝑑𝑒𝑚𝑎𝑠 𝑑𝑒 𝑙𝑎 𝑔𝑟𝑎𝑓𝑖𝑐𝑎 𝑑𝑒 𝑓 𝑠𝑒 𝑣𝑒 𝑞𝑢𝑒 𝑐𝑢𝑎𝑛𝑑𝑜 lim+ 𝑥→4

∴ lim 𝑥→ 4

3 3 = +⋈; 𝑦 𝑐𝑢𝑎𝑛𝑑𝑜 lim− = −⋈ 𝑥→4 𝑥−4 𝑥−4

3 =⋈ 𝑥−4

4. Dadas las funciones encuentra las asíntotas horizontales. 1) lim

𝑥→⋈

3𝑥 1+𝑥

𝑓(𝑥) = 𝑦 =

3𝑥 𝑦 → 𝑦(1 + 𝑥) = 3𝑥 → 𝑦 + 𝑦𝑥 = 3𝑥 → 𝑦 = 3𝑥 − 𝑦𝑥 → 𝑦 = 𝑥(3 − 𝑦) → 𝑥 = 1+𝑥 3−𝑦 3−𝑦 = 0 → 3 =𝑦 → 𝑦 =3 𝑇𝑖𝑒𝑛𝑒 𝑢𝑛𝑎 𝑎𝑠𝑖𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑛𝑧𝑜𝑛𝑡𝑎𝑙 𝑒𝑛 𝑦 = 3

𝐼𝑚 𝑓 = {𝑦0 ∈ ℝ | 𝑦0 ≠ 3 }

Hernández Martínez Marco Antonio

42800354

Página 13

CALCULO DIFERENCIAL Marco Antonio Hernández Martínez Maestría en Educación Matemática

lim

𝑥→⋈

3𝑥 1+𝑥

3𝑥 3𝑥 lim 3 3 3 3 𝑥 𝑥→⋈ lim = lim 𝑥 = lim = = = =3 𝑥 1 𝑥→⋈ 1 + 𝑥 𝑥→⋈ 1 𝑥→⋈ 1 0+1 1 + +1 lim + lim1 𝑥 𝑥 𝑥 𝑥 𝑥→⋈ 𝑥 𝑥→⋈ ∴ lim

3𝑥

𝑥→⋈ 1+𝑥

2) lim

𝑥→⋈

=3

4𝑥 − 3 5𝑥 + 4

𝑓(𝑥) = 𝑦 =

4𝑥 − 3 → 𝑦(5𝑥 + 4) = 4𝑥 − 3 → 5𝑥𝑦 + 4𝑦 = 4𝑥 − 3 → 4𝑦 + 3 = 4𝑥 − 5𝑥𝑦 5𝑥 + 4

4𝑦 + 3 = 𝑥(4 − 5𝑦) → 𝑥 =

4𝑦 + 3 4 − 5𝑦

4 − 5𝑦 = 0 → 4 = 5𝑦 → 𝑦 =

4 5

𝑇𝑖𝑒𝑛𝑒 𝑢𝑛𝑎 𝑎𝑠𝑖𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑛𝑧𝑜𝑛𝑡𝑎𝑙 𝑒𝑛 𝑦 =

Hernández Martínez Marco Antonio

42800354

4 5

Página 14

CALCULO DIFERENCIAL Marco Antonio Hernández Martínez Maestría en Educación Matemática

𝐼𝑚 𝑓 = {𝑦0 ∈ ℝ | 𝑦0 ≠

4 } 5

4𝑥 − 3 4𝑥 3 3 − 4− 4𝑥 − 3 𝑥 𝑥 𝑥 𝑥= lim = lim = lim = lim 4 4 𝑥→⋈ 5𝑥 + 4 𝑥→⋈ 5𝑥 + 4 𝑥→⋈ 5𝑥 𝑥→⋈ 5+ + 𝑥 𝑥 𝑥 𝑥 lim 4 − lim

𝑥→⋈

3

𝑥→⋈ 𝑥

4 lim 5 + lim 𝑥→⋈ 𝑥→⋈ 𝑥 ∴ lim

𝑥→⋈

4) lim

𝑥→⋈

=

4−0 4 = 5−0 5

4𝑥 − 3 4 = 5𝑥 + 4 5

4𝑥 𝑥+4 𝑓(𝑥) = 𝑦 =

4𝑥 → 𝑦(𝑥 + 4) = 4𝑥 → 𝑥𝑦 + 4𝑦 = 4𝑥 → 4𝑦 = 4𝑥 − 𝑥𝑦 → 4𝑦 = 𝑥(4 − 𝑦) 𝑥+4 4𝑦 4𝑦 =𝑥 →𝑥= (4 − 𝑦) (4 − 𝑦) 4−𝑦=0 →4=𝑦 →𝑦=4 𝑇𝑖𝑒𝑛𝑒 𝑢𝑛𝑎 𝑎𝑠𝑖𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑛𝑧𝑜𝑛𝑡𝑎𝑙 𝑒𝑛 𝑦 = 4

Hernández Martínez Marco Antonio

42800354

Página 15

CALCULO DIFERENCIAL Marco Antonio Hern├бndez Mart├нnez Maestr├нa en Educaci├│n Matem├бtica

ЁЭР╝ЁЭСЪ ЁЭСУ = {ЁЭСж0 тИИ тДЭ | ЁЭСж0 тЙа 4 }

lim

ЁЭСетЖТтЛИ

4ЁЭСе 4ЁЭСе lim 4 4ЁЭСе 4 ЁЭСе = lim 4 = ЁЭСетЖТтЛИ = lim ЁЭСе = lim = =4 4 4 ЁЭСетЖТтЛИ ЁЭСе + 4 ЁЭСетЖТтЛИ ЁЭСе + 4 ЁЭСетЖТтЛИ ЁЭСе + 4 1 + 0 1+ lim 1 + lim ЁЭСе ЁЭСе ЁЭСе ЁЭСе ЁЭСетЖТтЛИ ЁЭСетЖТтЛИ ЁЭСе тИ┤ lim

ЁЭСетЖТтЛИ

5) lim

ЁЭСетЖТтЛИ ЁЭСе 2

4ЁЭСе =4 ЁЭСе+4

ЁЭСе+3 тИТ 5ЁЭСе + 1

1 3 ЁЭСе+3 ЁЭСе 3 1 3 lim + lim 2 + 0+0 0 2 2 + ЁЭСе2 ЁЭСе ЁЭСе2 ЁЭСе ЁЭСе ЁЭСетЖТтЛИ ЁЭСе ЁЭСетЖТтЛИ ЁЭСе lim 2 = lim 2 = lim = = = =0 5 1 5 1 ЁЭСетЖТтЛИ ЁЭСе тИТ 5ЁЭСе + 1 ЁЭСетЖТтЛИ ЁЭСе ЁЭСетЖТтЛИ 5ЁЭСе 1 1 тИТ + 2 lim 1 тИТ lim + lim 2 1 + 0 + 0 1 тИТ + ЁЭСе ЁЭСе ЁЭСе2 ЁЭСе2 ЁЭСе2 ЁЭСе2 ЁЭСетЖТтЛИ ЁЭСетЖТтЛИ ЁЭСе ЁЭСетЖТтЛИ ЁЭСе

тИ┤ lim

ЁЭСетЖТтЛИ

ЁЭСУ(ЁЭСе) =

Hern├бndez Mart├нnez Marco Antonio

ЁЭСе2

ЁЭСе+3 =0 ЁЭСе 2 тИТ 5ЁЭСе + 1

ЁЭСе+3 ЁЭСбЁЭСЦЁЭСТЁЭСЫЁЭСТ ЁЭСвЁЭСЫЁЭСО ЁЭСОЁЭСаЁЭСЦЁЭСЫЁЭСбЁЭСЬЁЭСбЁЭСО тДОЁЭСЬЁЭСЯЁЭСЦЁЭСзЁЭСЬЁЭСЫЁЭСбЁЭСОЁЭСЩ ЁЭСТЁЭСЫ ЁЭСж = 0 тИТ 5ЁЭСе + 1

42800354

P├бgina 16

CALCULO DIFERENCIAL Marco Antonio Hernández Martínez Maestría en Educación Matemática

7) lim

𝑥→⋈ 𝑥 2

1−𝑥 + 4𝑥 + 4

1−𝑥 1 𝑥 1 1 1−𝑥 2 2 − 𝑥2 2−𝑥 𝑥 𝑥 𝑥 lim 2 = lim 2 = lim 2 = lim = 4 4 𝑥→⋈ 𝑥 + 4𝑥 + 4 𝑥→⋈ 𝑥 + 4𝑥 + 4 𝑥→⋈ 𝑥 𝑥→⋈ 4𝑥 4 1 + + + + 2 𝑥 𝑥 𝑥2 𝑥2 𝑥2 𝑥2

lim

1

𝑥→⋈ 𝑥

=

2

− lim

lim 1 + lim

𝑥→⋈

4

𝑥→⋈ 𝑥

+ lim

∴ lim

𝑥→⋈

𝑓(𝑥) =

Hernández Martínez Marco Antonio

𝑥2

1

𝑥→⋈ 𝑥

4

𝑥→⋈ 𝑥

2

=

0−0 0 = =0 1+0+0 1

1−𝑥 =0 𝑥 2 + 4𝑥 + 4

1−𝑥 𝑡𝑖𝑒𝑛𝑒 𝑢𝑛𝑎 𝑎𝑠𝑖𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑛𝑧𝑜𝑛𝑡𝑎𝑙 𝑒𝑛 𝑦 = 0 + 4𝑥 + 4

42800354

Página 17

CALCULO DIFERENCIAL Marco Antonio Hernández Martínez Maestría en Educación Matemática

8) lim

𝑥→⋈

2𝑥 3 + 7 5𝑥 2 − 1

7 2𝑥 3 + 7 2𝑥 3 7 7 lim 2 + lim 3 + 3 2+ 3 2𝑥 3 + 7 3 2+0 𝑥→⋈ 𝑥 𝑥 𝑥3 𝑥 = lim 𝑥 = 𝑥→⋈ lim = lim = lim = = 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑑𝑎 1 5 1 𝑥→⋈ 5𝑥 2 − 1 𝑥→⋈ 5𝑥 2 − 1 𝑥→⋈ 5𝑥 2 𝑥→⋈ 5 1 0 − lim − lim − 𝑥 𝑥3 𝑥→⋈ 𝑥 𝑥→⋈ 𝑥 3 𝑥3 𝑥3 𝑥3

𝑓(𝑥) =

Hernández Martínez Marco Antonio

2𝑥 3 + 7 𝑛𝑜 𝑡𝑖𝑒𝑛𝑒 𝑎𝑠𝑖𝑛𝑡𝑜𝑡𝑎𝑠 ℎ𝑜𝑟𝑖𝑛𝑧𝑜𝑛𝑡𝑎𝑙𝑒𝑠 5𝑥 2 − 1

42800354

Página 18

Turn static files into dynamic content formats.

Create a flipbook
Issuu converts static files into: digital portfolios, online yearbooks, online catalogs, digital photo albums and more. Sign up and create your flipbook.