Investigacion limites

Investigación

Límites de una función

Clase 9 de febrero de 2014

Tabular el siguiente límite

1. Lím x > 2

5x²-3x+1/ x+1        

5(-1) ² - 3(-1) ² +1/ -1+1 = 5+3+1/0 = 0

5 (-0.1) ² -3(-0.1) ²+1/ -0.1+1= 0.05-0.03+1/0.9 = 1.02/0.9 = 1.133

5(-0.01) ²-3(-0.01) ²+1/ -0.01+1 = 0.0005-0.0003+1/0.99 = 1.0002/0.99 = 1.01

5(-0.001) ²-3(-0.001) ²+1 / -0.001+1 =  0.000005-0.000003+1 / 0.999 = 1.000002/0.999 = 1.001

5(-0.0001) ²-3(-0.0001) ²+1 / -0.00001+1= 0.00000005 – 0.00000003 +1/ -0.0001+1 = 1.0001

5(0) ²-3(0) ²+1/ 0 +1 = 1/1 = 1

5(1) ²-3(1) ²+1/ 1 +1 = 5-3+1/0= 1.5

5(1.09) ²-3(1.09) ²+1/ 1.09 +1 = 5.9-3.5+1 / 2.9= 1.17

5(1.9) ²-3(1.9) ²+1/ 1.9 +1 = 18.05-10.83+1 / 2.9= 2.83

5(2) ²-3(2) ²+1/ 2 +1 = 20-6+1/3 = 15/3 = 5

5(2.01) ²-3(2.01) ²+1/ 2.01 +1 = 20.2-12.12+1 / 3.01 = 3.01

Factorizar o/y sustituir los siguientes límites

1. Lim x->-6

3x ²+7x -2 =

3(-6)²+7(-6)-2 = 108-42-2 = 64

1.  Lim x->2

√ (x ²-3x -6) =

[√ (2 ²-3(2) -6) ] ² = 8-36-36= 8

1. Lím x->3

[3x²-26)] ^27 =

[3(3) ²-26]^27  = [27-26]^27 = 1^27 = 1

1. Lim x-> 3

(2x²+7x+10/x-2) ^1/2

(2(3)²+7(3)+10/3-2) ^1/2

(18+21+20/1) ^1/2 = (49) ^1/2  = 24.5

1. Lím x -> 3

x^3-2x²-9/x-3

3^3  - 2(3) ² -9 / 3-3 =    27-18-9/0  = 0/0 = 0

1. Lim x -> -1

2x² - 1 / x² -1 = 1

1. Lím x -> 3

2x² -5x-3/ x² +2x -15 =   2(3) ²-5(3)-3 / (3) ²+ 2(3) -15 = 18-15-3/ 6+6-15 = 0/0 = 0

1. Lim y -> -4

y²+9y+20/ y²+y -12  =   (y+5)(y+4) / (y+4)(y-3)

(y+5)/(y-3) = -4+5/-4-3  = -1/7

1. Lím

3w² +16w-35 / w²+4w-21 =

(3w-5)(w+7) / (w+7)(w-3) =

3w-5/w-3 =     3(-4)+5/-4-3 =   -12-5/-7 =   -17/7

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