Calculo 1 - sem 6 - Unmsm 2020 PDF

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UNIVERSIDAD NACIONAL MAYOR DE SAN MARCOS

ESCUELA DE ESTUDIOS GENERALES

Estudiante:

Diciembr, 2021

lim 𝑥→ ∞

lim 𝑥→∞

3 𝑥 +4𝑥 3

2𝑥 +3𝑥 +4 3

2𝑥 +3𝑥 +4 3

lim 𝑥→∞

2

+

4 𝑥 𝑥

2

3

2𝑥 +3𝑥 +4 3

2

3+4

=

2+

3

3.1 𝑥

1 𝑥

+

4

𝑥

3

=

3

2

𝑥 +𝑥+3 (𝑥−1)(𝑥+1) 2

𝑥 +𝑥+3 (𝑥−1)(𝑥+1) 1+

1 𝑥

1−

lim

𝑥

3

3

𝑥

2

𝑛→∞

3 𝑥

=

3 𝑥 +4𝑥 3

lim 𝑥→∞

=

2

+ 1

𝑥

3

𝑥

2

2

= lim 𝑥→ ∞

=

𝑛 +𝑛+1 2

𝑛 +2𝑛+1 3

1+0+0 1−0

𝑥 +𝑥+3 2

𝑥 −1

= 1

2

=

𝑥

𝑥

2

2

+ 𝑥 𝑥

2 2

𝑥

𝑥

2

−

+ 1

𝑥

2

3

𝑥

2

=

1+

1 𝑥

1−

+

1

𝑥

2

3

𝑥

2

lim

2

∞

3

lim 𝑥→ ∞

lim 𝑥→∞ =

4

4

4

2

lim = →

𝑛 +𝑛+1

𝑛 +2𝑛+1 → ∞+0+0 = 1+0+0 = ∞

𝑛

𝑛 +𝑛+1

𝑛

∞

𝑥 −3 +

𝑛 +2𝑛+1 3

𝑛 3 𝑛

4

=

𝑛

3

𝑛

3

𝑛

3

𝑛

3

+ 3

+

𝑛

2

𝑛

+

2𝑛 𝑛

3

+

1

2

1

2

𝑛

3

𝑛

3

𝑛+

=

1+

1 𝑛

2

𝑛

2

+

1

𝑛

3

+

1

𝑛

3

𝑥 +5𝑥+6

4

2

3𝑥+8− 8𝑥 +9𝑥+8

𝑥 −3 +

2

𝑥 +5𝑥+6

4

2

3𝑥+8− 8𝑥 +9𝑥+8 2

1− 0+ 1+0+0

3+0− 8+0+0

=

4

= lim 𝑥→ ∞

1+ 1

3− 8

=

4

4

2

+

𝑥 +5𝑥+6

+𝑥 −

8𝑥 +9𝑥+8

𝑥 −3

3𝑥 𝑥

3−2 2

4

𝑥

.

4

8

2

2

𝑥

2

𝑥

2

(3+2 2)

(3+2 2)

=

4

𝑥

𝑥

4 4

3𝑥 𝑥

−

3

𝑥

4

+𝑥− 8

=6+4 2

+

2

+

5𝑥

+

9𝑥

+

𝑥

8𝑥 𝑥

2

2

𝑥

2

𝑥

2

𝑥

+

2

8

𝑥

6

𝑥

2

2

lim ( 𝑥 + 2𝑥 − 𝑥) 𝑥→ ∞ 2

lim ( 𝑥 + 2𝑥 − 𝑥) = lim ( 𝑥 + 2𝑥 − 𝑥) · 𝑥→∞ 𝑥 →∞ 2

=

∞+2+1

1+0+1

2

=

∞ 2

= ∞

lim ( 𝑥 + 2𝑥 − 𝑥 →∞

𝑥 − 2𝑥)

( 𝑥 +2𝑥+𝑥) 2

( 𝑥 +2𝑥+𝑥) 2

=

𝑥 +2𝑥−𝑥 2

2

𝑥 +2𝑥+𝑥 2

=

𝑥 𝑥

2

+

2𝑥 𝑥 +𝑥 𝑥

𝑥 +2𝑥 2

𝑥

2

+

𝑥 𝑥

=

lim∞

𝑥

→ lim

𝑥

→

𝑥+ 2𝑥−𝑥+ 2𝑥

∞

𝑥+ 2𝑥+ 𝑥− 2𝑥 𝑥

= 2 2 · lim 𝑥 →∞

lim → 𝑥 −∞ lim 𝑥 → −∞

3𝑥+ 𝑥 +9𝑥+8 2

=

3𝑥+ 𝑥 +9𝑥+8

5𝑥 + 𝑥

9𝑥

−

𝑥

3𝑥 𝑥

𝑥

𝑥

2

2

+ +

7𝑥 𝑥

2

9𝑥 𝑥

2

+

+

8

𝑥

=

2

8

𝑥

2

5+

9+

3−

1+

7 𝑥

9 𝑥

+ +

8

𝑥

2

𝑥

2

=

8

5+ 9

3− 1

=

8 2

=4

2

4𝑥− 27𝑥 +16𝑥 +4𝑥+1 3

3

2

(

2

4𝑥− 27𝑥 +16𝑥 +4𝑥+1 3

3

2

𝑥 𝑥+1

−

−

3𝑥 +1 3𝑥−2

2

𝑥 𝑥+1 2

(

2

2

3𝑥+ 𝑥 −6𝑥+8

3𝑥+ 𝑥 −6𝑥+8

lim 𝑥 → +∞

= 2

1 2

2

2

(

𝑥− 2𝑥

𝑥+ 2𝑥

5𝑥− 9𝑥 +7𝑥+8

lim 𝑥 → +∞ lim 𝑥 → +∞

𝑥− 2𝑥

2

lim 𝑥 → −∞

= 2 2·

𝑥+ 2𝑥

𝑥+ 2𝑥+ 𝑥− 2𝑥

5𝑥− 9𝑥 +7𝑥+8

lim 𝑥 → −∞

=

lim 𝑥 − 2𝑥) = ( 𝑥 + 2𝑥 − 𝑥 − 2𝑥) · 𝑥 →∞ lim lim 2 2𝑥 𝑥 = = 2 2· 𝑥+ 2𝑥+ 𝑥+ 2𝑥+ 𝑥− 2𝑥 𝑥→ ∞ 𝑥→∞

( 𝑥 + 2𝑥 −

𝑥 (𝑥−1) 2

𝑥 −1 2

2

)−

3𝑥 +1 3𝑥−2 2

)

=

)

3𝑥 − 𝑥 4𝑥 𝑥

27𝑥 𝑥

3 𝑥

3

3

( )

( )

= lim 𝑥 → +∞

lim 𝑥 → +∞

+

3

𝑥

𝑥 𝑥+1 2

+

3−

1 𝑥

2 𝑥

=

𝑥

2

2

+

−

6𝑥 𝑥

16𝑥

−

𝑥

3

2

2

+

+

8

𝑥

2

4𝑥 𝑥

3

+

lim 𝑥 → +∞

lim 𝑥 → +∞

1

𝑥

=

3

(

3− 4𝑥 𝑥

3𝑥 +1 3𝑥−2 2

)

+

=

( )−( 𝑥 −𝑥 3

𝑥 −1 2

3

1− 𝑥 + 6

27+

16 𝑥

+

lim 𝑥 →+∞

0+0 3−0

)=

8

𝑥

2

4

𝑥

2

(

+

1

𝑥

=

3

𝑥 𝑥+1 2

lim 𝑥 → +∞

3− 1−0+0

4+ 27+0+0+0 3

)−

( )

·

𝑥−1 𝑥−1

𝑥 − 2

1−

1 𝑥

1

𝑥

2

−

lim

𝑥 → +∞

=

(

( ) 0+0 3−0

=

𝑥

𝑥

lim

→

→

+∞

lim

lim 𝑥 → +∞ =

2

+∞

(

(

(

2𝑥 +3 2

𝑥

2

2

𝑥

− 𝑥

2

𝑥 −1 2

𝑥

𝑥 −1 2

lim ⎛ 𝑥 → +∞ ⎝

lim 𝑥→∞

𝑥

2

lim 𝑥→ ∞

−0=1−0=1

𝑥 11

( ) 1− 𝑥 −

(

)

𝑥

−𝑥 ·

(

𝑥 −1 𝑥

2

𝑥

2

2𝑥 + 3 − 2𝑥

(

2𝑥 −5 2

𝑥

2

lim 𝑥 𝑥→∞

(

=

𝑥 −𝑥

𝑥 + 1−

3

𝑥 +𝑥 + 𝑥 +𝑥 4

2

6

3

=

=

2+0+ 2−0

2

5

3

( − 5) (

2

lim 𝑥 𝑥 + 1 − 𝑥 𝑥 𝑥→∞ 4

2

0

2

𝑥

𝑥

𝑥 +𝑥 4

𝑥

4

4

2

2

+

)

2

⎞ = lim ⎛ +𝑥 𝑥 → +∞ 𝑥 −1 ⎠ ⎝ 𝑥

2

)

2

𝑥

𝑥 (𝑥+𝑥) 2

2

3

2

2 2

=0

0

=

lim 𝑥 → +∞

=

2𝑥 +3+ 2𝑥 −5 2

( ) 𝑥

𝑥

2

𝑥 −𝑥 +𝑥 4

𝑥

2

2𝑥 +3+ 2𝑥 −5 2

2

=

2

2

2

𝑥 −1+𝑥(𝑥 −1)

2

2𝑥 +3−(2𝑥 −5)

4

𝑥 −1 2

𝑥 −1 2

1 𝑥

=

2

=0

⎞

⎠

8

=

2𝑥 +3+ 2𝑥 −5 2

2

2𝑥

)

3

5

𝑥

) )

2𝑥 +3+ 2𝑥 −5

2 2

( + 1) (

𝑥

(

−𝑥

𝑥 + 1

3

3

−

lim 𝑥 → +∞

4

𝑥 −1 2

2𝑥 − 5

2

2𝑥 + 3 −

+

)⎠

⎞=

𝑥 −1 𝑥+ 𝑥 −1 2

𝑥

= lim ⎛ 𝑥 → +∞ ⎝

+𝑥

2

𝑥 −1

2

8 𝑥

)

+𝑥

2

2

𝑥 +𝑥 6

𝑥

6

3

) )

𝑥 𝑥 +1+𝑥 𝑥 +1 2

3

3

=

𝑥 𝑥 +1+𝑥 𝑥 +1

=

2

3

𝑥 −𝑥 2

1+

3

2 +

1

𝑥

3

3

1+

1

𝑥

3

𝑥 (𝑥 +1)−𝑥 (𝑥 +1)

=

2

2

2

3

𝑥 𝑥 +1+𝑥 𝑥 +1 2

−∞

1+ 1

3

3

=− ∞

=

𝑥 −𝑥 4

𝑥

2

5

𝑥 +1 + 𝑥 2

3

3

3

𝑥 +1 3

:

lim

𝑥

𝑥 +2𝑥 2

2 → 𝑥 +2𝑥

∞ 4𝑥2−1

=

4 1

2 1 ⎡(𝑠𝑒𝑎 ε > 0), (∃ 𝑀 > 0)/(𝑠𝑖 𝑥 > 𝑀 ∧ 𝑥 ∈ 𝐷𝑜𝑚𝑓) → || 𝑥 +2𝑥 − lim ↔ = ⎢ 2 4 | 4𝑥2−1 ⎣ 𝑥 → ∞ 4𝑥 −1 | 4𝑥2+8𝑥−4𝑥2+1 | | 8𝑥+1 | | 𝑥2+2𝑥 1 | | = | 2 | 0), (∃ 𝑀 > 0)/( 𝑠𝑖 𝑥 > 𝑀 ∧ 𝑥 ∈ 𝐷𝑜𝑚𝑓) → | 𝑥 + 6𝑥 + 1 | ⎣ 𝑥→ ∞

3 2 |3 3 | 𝑥 + 6𝑥2 + 1 − (𝑥 + 2) · (𝑥 +6𝑥 +1 ) | | ( 𝑥3+6𝑥2+1) 3 3 2 3 | | 𝑥 +6𝑥 +1 −( 𝑥+2) | | =| | | ( 𝑥3+6𝑥2+1) +( 𝑥3+6𝑥2+1) (𝑥+2)+(𝑥+2)2 | | |

(

=

)

3

2 3

−12𝑥−7

2

lim 𝑥 → −∞

lim 𝑥 → −∞

4−𝑥

𝑥 +𝑥−1 2

3

4−𝑥

2

𝑥 +𝑥−1 2

2 3

(3 2 ) 3 2 +(𝑥 +6𝑥 +1 ) + 𝑥 +6𝑥 +1

1 3 1 3

( 𝑥+2)+( 𝑥+2)

2

( 𝑥+2)+( 𝑥+2)

2

| | | |

1 3

(𝑥 +6𝑥 +1 ) + (𝑥 +6𝑥 +1 ) 3

2 3

2 3

1 3

( 𝑥+2)+( 𝑥+2)

= 0

2

< ε

(

| 4−𝑥 | = 0 ↔ ⎡⎢ (𝑠𝑒𝑎 ε > 0), (∃ 𝑀 < 0)/(𝑠𝑖 𝑥 < 𝑀 ∧ 𝑥 ∈ 𝐷𝑜𝑚𝑓) → | 2 − 0| < | | 𝑥 +𝑥−1 ⎣

| | |

| | | 𝑥2+𝑥−1 | < ε | | 4−𝑥

→

4−𝑥

<

𝑥 +𝑥−1 2

lim 𝑥 → −∞

lim 𝑥 → −∞

→

4

𝑥 −1 2

= 0

400

𝑥 +1 2

< ε → 4 < ε𝑥 − ε → 2

4

𝑥 −1 2

𝑥 +1 2

(

)

2 | 400 | 2 | | < ε → 400 < ε 𝑥 + 1 → 400 < ε𝑥 + ε → | 𝑥2+1 |

lim

lim − 𝑥→3

−

lim

𝑡 → −3

lim + 𝑡 → −3

2

+

lim 3

lim

𝑥→ 5 3

5−𝑥

3

)

400−ε ε

0} = (-4)>0 Do(F)= 𝑈 < 4, + ∞ > 2

F(-X)=

( (

−3𝑥−1

𝑥 +4𝑥 2

) )

𝑥 − 4𝑥 = 0 → 𝑥 − 4𝑥 = 0 → 𝑥(𝑥 − 4) = 0 → 𝑥 = 0 ∧ 𝑥 = 4 2

lim 𝑥→ 0

lim 𝑥→ 4

lim

𝑥 → +∞

3𝑥−1

𝑥 −4𝑥 2

3𝑥−1

𝑥 −4𝑥

(

2

3𝑥−1

2

= =

𝑥 −4𝑥 2

)

3(0)−1

=

3 0

3(4)−1

=

11 0

(0) −4(0) 2

(4) −4(4) 2

= lim 𝑥 → +∞

Por end →+∞ →-∞

𝑁𝑜 𝑒𝑥𝑖𝑠𝑡𝑒

= ∞, 0 𝑒𝑠 𝑎. 𝑣.

𝑥

2

2

= ∞, 4 𝑒𝑠 𝑎. 𝑣

( ) 3𝑥−1

1

= lim 𝑥 → +∞

(

3𝑥−1 𝑥

)=

lim

𝑥→ +∞

( ) 3−

1

1 𝑥

= 3, 3 𝑒𝑠 𝑎. ℎ.

𝑓(𝑥) =

( (

3𝑥 −12 2

𝑥 −6𝑥 2

) )=

𝑥 − 6𝑥 = 0 → 𝑥(𝑥 − 6) = 0 → 𝑥1 = 0, 𝑥2 = 6 2

lim 𝑥→ 0

lim 𝑥→ 6

lim 𝑥 → +∞

3𝑥 −12 2

𝑥 −6𝑥 2

3𝑥 −12 2

𝑥 −6𝑥

(

2

𝑁𝑜 𝑒𝑥𝑖𝑠𝑡𝑒.

3𝑥 −12 2

𝑥 −6𝑥 2

=

)

3(0) −12 2

=

−12 0

3(6) −12

=

96 0

(0) −6(0) 2

2

(6) −6(6)

=

2

lim 𝑥 → +∞

= ∞, 6 𝑒𝑠 𝑎. 𝑣.

( ) 3−

12

𝑥 6 1− 𝑥

2

= ∞, 0 𝑒𝑠 𝑎. 𝑣.

= 3, 3 𝑒𝑠 𝑎. ℎ.

𝑓(𝑥) =

( lim ( →

3𝑥 −1 3

𝑥 −2𝑥 2

)= )=

𝑥 − 2𝑥 = 0 → 𝑥(𝑥 − 2) = 0 → 𝑥1 = 0, 𝑥2 = 2 2

lim 𝑥→ 0 𝑥

2

lim 𝑥 → +∞ 3𝑥 −1 3

𝑥 −2𝑥 2

3𝑥 −1 3

𝑥 −2𝑥 2

3𝑥 −1

(

3

𝑥 −2𝑥 2

3𝑥 −1 3

𝑥 −2𝑥 2

)

3(0) −1

=

−1 0

= ∞, 0 𝑒𝑠 𝑎. 𝑣.

3(2) −1

=

23 0

= ∞, 2 𝑒𝑠 𝑎. 𝑣.

3

(0) −2(0) 2

3

(2) −2(2) 2

=

lim 𝑥 → +∞

( ) 3𝑥− 1−

1

𝑥 2 𝑥

2

= 3𝑥 + 6, 3𝑥 + 6 𝑒𝑠 𝑎. 𝑜.

= ∞, 𝑛𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑎. ℎ.

𝑓(𝑥) =

(

𝑥 +2𝑥 3

𝑥 −2𝑥−3 2

)

𝑥 − 2𝑥 − 3 = 0 → (𝑥 − 3)(𝑥 + 1) = 0 → 𝑥1 = 3, 𝑥2 =− 1 2

lim 𝑥→ 1

lim 𝑥 → −1

lim 𝑥 → +∞ 𝑥 +2𝑥 3

(

3

𝑥 −2𝑥−3

(

𝑥 −2𝑥−3 2

𝑥 +2𝑥

2

𝑥 +2𝑥 3

𝑥 −2𝑥−3 2

𝑥 +2𝑥 3

𝑥 −2𝑥−3 2

)

=

(1) +2(1)

(1) −2(1)−3 2

=

)=

3

(

=

3 −4

3

(−1) −2(−1)−3 2

lim 𝑥 → +∞

𝑥+ 𝑥

2

1− 𝑥 − 2

= 𝑥 + 2, 𝑥 + 2 𝑒𝑠 𝑎. 𝑜.

)

=

(−1) +2(−1)

3

𝑥

2

, 1 𝑛𝑜 𝑒𝑠 𝑎. 𝑣. −3 0

= ∞, − 1 𝑒𝑠 𝑎. 𝑣.

= ∞, 𝑛𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑎. ℎ.

𝑓(𝑥) =

( (

5𝑥 +1 2

𝑥 −2𝑥 2

)= )=

𝑥 − 2𝑥 = 0 → 𝑥(𝑥 − 2) = 0 → 𝑥1 = 0 ∧ 𝑥2 = 2 2

lim 𝑥→ 0

lim 𝑥→ 2

lim 𝑥 → +∞

5𝑥 +1 2

𝑥 −2𝑥 2

5𝑥 +1 2

𝑥 −2𝑥

(

2

5𝑥 +1 2

𝑥 −2𝑥

𝑁𝑜 𝑒𝑥𝑖𝑠𝑡𝑒.

2

)

=

5(0) +1 2

(0) −2(0) 2

(2) −2(2) 2

lim 𝑥 → +∞

1 0

=

21 0

( )

5(2) +1 2

=

5+

1

𝑥 2 1− 𝑥 2

= ∞, 0 𝑒𝑠 𝑎. 𝑣. = ∞, 2 𝑒𝑠 𝑎. 𝑣.

= 5, 5 𝑒𝑠 𝑎. ℎ.

𝑓(𝑥) =

(

𝑥

3

(𝑥−3)

2

)=

𝑥 − 3 = 0 ∧− (𝑥 − 3) = 0 → 𝑥 − 3 = 0 ∧− 𝑥 + 3 = 0 → 𝑥 = 3

lim 𝑥→ 3

lim 𝑥 → +∞ 𝑥

3

(𝑥−3)

2

𝑥

3

(𝑥−3)

(

2

𝑥

3

(𝑥−3)

2

(3)

3

(3−3)

)=

2

=

lim 𝑥 → +∞

(

27 0

= ∞, 3 𝑒𝑠 𝑎. 𝑣.

𝑥

3

𝑥 −6𝑥+9 2

= 𝑥 + 6, 𝑥 + 6 𝑒𝑠 𝑎. 𝑜.

)=

lim 𝑥 → +∞

(

𝑥

1− + 6 𝑥

9

𝑥

2

)

= ∞, 𝑛𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑎

𝑓(𝑥) =

(

3𝑥 −12𝑥+12 2𝑥−2 2

)

2𝑥 − 2 = 0 → 2𝑥 = 2 → 𝑥 = 1

lim 𝑥→ 1

lim 𝑥 → +∞

3𝑥 −12𝑥+12 2𝑥−2 2

(

3𝑥 −12𝑥+12 2𝑥−2 2

3𝑥 −12𝑥+12 2𝑥−2 2

=

3 2

=

)=

𝑥−

=

3(1) −12(1)+12 2(1)−2 2

lim 𝑥 → +∞

9 2

(

3𝑥−12+ 2−

2 𝑥

12 𝑥

3 0

)

= ∞, 1 𝑒𝑠 𝑎. 𝑣. = ∞, 𝑛𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑎. ℎ.

𝑓(𝑥) =

3𝑥 −2𝑥 5

𝑥 −13𝑥 +36 4

2

(

)( 2

)

𝑥 − 13𝑥 + 36 = 0 → 𝑥 − 4 𝑥 − 9 = 0 → 𝑥1 = 2 ∧ 𝑥2 =− 2 ∧ 𝑥3 = 3 ∧ 𝑥4 = 4

lim 𝑥→ 2

(

lim 𝑥→ 3

(

2

lim 𝑥 → −2

(

lim 𝑥 → −3

(

lim

𝑥 → +∞

5

𝑥 −13𝑥 +36 4

2

3𝑥 −2𝑥 5

2

3𝑥 −2𝑥 5

𝑥 −13𝑥 +36 4

)= )= )= )=

(

2

3𝑥 −2𝑥 5

𝑥 −13𝑥 +36 4

2

3𝑥 −2𝑥 5

𝑥 −13𝑥 +36 4

𝑥 −13𝑥 +36 4

2