Examen de Matematicas 2o de Bachillerato CSOctubre 2011

Problema 1 Calcular los siguientes lmites:

1. lmx

4x4 3x3 x 73x4 + 5x 1

2. lmx

x2 + 5x 77x3 + x2 x + 1

3. lmx

6x4 + 5x3 x 17x2 2

4. lmx

(3x2 + 8x 122x2 + 5x 1

)9x+5

5. lmx

(7x2 6x 310x2 x + 2

) 3x+42

6. lmx

(3x 93x 1

)x+2Solucion:

1. lmx

4x4 3x3 x 73x4 + 5x 1 =

4

3

2. lmx

x2 + 5x 77x3 + x2 x + 1 = 0

3. lmx

6x4 + 5x3 x 17x2 2 =

4. lmx

(3x2 + 8x 122x2 + 5x 1

)9x+5=

5. lmx

(7x2 6x 310x2 x + 2

) 3x+42

= 0

6. lmx

(3x 93x 1

)x+2= e8/3

Problema 2 Calcular los siguientes lmites:

1. lmx

3x2 x 1

3x + 2

1

2. lmx

2x5 + 58x + 5

3. lmx

3x2 x + 12x2 8x + 3

4. lmx

5x3 + 2x2 1x2 2

5. lmx

(2x2 x + 2

2x2 + 5x 1

)6. Sabiendo que lm

x

(5x + 1

5x 1)2nx

= 3, calcular n.

Solucion:

1. lmx

3x2 x 1

3x + 2=

3

3

2. lmx

2x5 + 58x + 5

=

3. lmx

3x2 x + 12x8x + 3

=

3

2

4. lmx

5x3 + 2x2 1x2 2 = 0

5. lmx

(2x2 x + 2

2x2 + 5x 1

)= 3

2

2

6. lmx

(5x + 1

5x 1)2nx

= 3 = n = 5 ln 34

= 1, 373265360.

Problema 3 Calcular los siguientes lmites:

1. lmx 2

3x2 5x 2x3 + x2 12

2. lmx 1

x4 + 3x3 3x 1x4 1

3. lmx 2

5x2 47x + 2

x 2Solucion:

1. lmx 2

3x2 5x 2x3 + x2 12 =

7

16

2

2. lmx 1

x4 + 3x3 3x 1x4 1 =

5

2

3. lmx 2

5x2 47x + 2

x 2 =13

8

3