# Limites

Páginas: 2 (403 palabras) Publicado: 9 de febrero de 2011
a) Obtén lim (+7x-5) x 3

Lim ( +7x-5) = lim +lim 7x-lim5
x 3 x 3 x 3

= (3+7(3)-5
=9+21-5
=25 (T.L.10)

b) Lim Lim Lim

x2 x2 x2Lim = (T.L.9)

Lim (T.L.4)

Lim (T.L.5)

Lim (T.L.1, T.L.2,T.L.3)

==

c) Lim

= lim

Lim (x+5) x->5 Lim x + Lim 5
x->5
=(5)+5=10

d) Comprueba que le limite (=6 x->2
Aplica los teoremas y especifica que teorema es.

Lim =6 lim + lim 3x – lim4
((2+3(2)-4)=6 x->2 x->2
(4+6-4)=6 (T.L.8)e)-Si f(x)= determina el valor del limite f(x) cuando x->
Lim = Lim (T.L.9)
= (T.L.5)
= (T.L.1, 2, 3)
==
f)-F(x)= encontrar el valor de Lim f(x) x->2
Lim= Lim= = (T.L.2, 8, 9)
g)-Obtén Lim x->3

Lim (T.L.9)
Lim (T.L. 4)
== (T.L.1,2,3)

Demuestra los siguientes límites

1.-Lim (7-2x) sol 3x->2

Lim7-lim 2x (T.L.1, 2)
x->2 x->2

7-2(2)=
7-4=3
2.-Lim (4 sol= 24
x->3

Lim 4-lim 2x-lim 6 (T.L.1)
x->3 x->3
4lim-2lim-lim6 (T.L.2)
x->3x->3
4lim (-2limx-lim6 (T.L.8)
x->3 x->3
4(9)-2(3)-6
36-6-6=24

+3.-Lim (6-3x) sol=18 x->-4

Lim 6-lim 3x (T.L. 1,2)
x->-4 x->-4

6-3(-4)
6-(-12)=184.-Lim sol= 0 r= 2

= 0

5.-Lim sol=7 Z=-2

lim

lim

lim
lim

lim

lim

lim

lim=7

6.-lim (x2-8) (4x-8) sol=64.X=4

Lim (lim x2-lim8) (lim 4x-lim8)
Lim (lim x2-lim8) (4lim x-lim8)
Lim (lim(x) 2-lim8) (4lim x-lim8)

((4)2-8) (4(4)-8)
(16-8) (16-8)
(8) (8)= 64

7.-lim (x2+) (x -) sol= -
X=-

Lim x3 -x2 + x -

Lim x3- limx2 + lim x – lim

Lim x3 - limx2 + lim x – lim

Lim ( )3 – () ()2 + (- ) -

Lim (-) – () () + (-) -

Lim - - - -

Lim= -

8.-Lim ( + ) (r2 - )...

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