Electronica Basica

Nota: Entrega presencial o en físico

Determine el comportamiento de las siguientes sucesiones, utilizando el concepto de límite y como dominio los cuatro primeros números naturales.

a)

Lim (-1)n-12n

n→1

= (-1)1-1 *21

= (-1)0 *2 = 1*2= 2

Lim (-1) n-12n

n→2

= (-1)n-1 *2n

= (-1)2-1 *22 = (-1) *4 = -4

Lim (-1) n-12n

n→3

= (-1) n-1 * 2n

= (-1)3-1 *23 = (-1)2 * 8 = 8

Lim (-1) n-12n

n→4

= (-1)n-1 * 2n

= (-1)4-3 *24 = (-1)3 *16 = -16

b)

Lim n+2

n→1 2n-1

= 1+2 = 3 = 3

2(1)-1 2-1

Lim n+2

n→2 2n-1

= 2+2 = 4 = 4

2(2)-1 4-1 3

Lim n+2

n→3 2n-1

= 3+2 = 5 = 1

2(3)-1 6-1

Lim n+2

n→4 2n-1

= 4+2 = 6 = 6

2(4)-1 8-1 7

c) an= n+1

n

Lim n+1

n→1 n

= 1+1 = 2

1

Lim n+1

n→2 n

= 2+1 = 3

2 2

Lim n+1

n→3 n

= 3+1 = 4

3 3

Lim n+1

n→4 n

= 4+1 = 5

4 4

d) an = 7n-2_____

n(n+2)(n+1

Lim 7n-2_______

n→1 n ((n+2) (n+1)

= 7(1)-2_______ = 5

1 (1+2) (1+1) 6

Lim 7n-2_______

n→2 n ((n+2) (n+1)

= 7(2)-2_______ = 12 = 1

2(2+2) (2+1) 24 2

Lim 7n-2_______

n→3 n ((n+2) (n+1)

= 7(3)-2_______ = 19

3(3+2) (3+1) 60

Lim 7n-2_______

n→4 n ((n+2) (n+1)

= 7(4)-2_______ = 26 = 13

4(4+2) (4+1) 120 60

e) an =3n(n+2)(n+1)

Lim 3n (n+2) (n+1)

n→1

= 3(1) (1+2) (1+1) 3(5) = 15

Lim 3n (n+2) (n+1)

n→2

= 3(2) (2+2) (2+1) 6(7) = 42

Lim 3n (n+2) (n+1)

n→3

= 3(3) (3+2) (3+1) 9(9) = 81

Lim 3n (n+2) (n+1)

n→4

= 3(4) (4+2) (4+1) 12(10) = 120

an= __ n+2___

2n(n+2) (n+1

Lim n+2_______

n→1 2n (n+2) (n+1)

= 1+2_______ = 3

2(1) (1+2) (1+1) 10

Lim n+2_______

n→2 2n (n+2) (n+1)

= 2+2_______ = 4_ = 2_

2(2) (2+2) (2+1) 28 14

Lim n+2_______

n→3 2n (n+2) (n+1)

= 3+2_______ = 5_

2(3) (3+2) (3+1) 54

Lim n+2_______

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