Question

calcule os limites:
lim x>0 (raiz de 1+x) - (raiz de 1-x)/x

lim x>1 3 -(raiz 10-x)/x^2-1


lim x>1 x^3-3x^2+6x-4/x^3-4x^2+8x-5

Answer 1

1)

No limite

lim x ->0 Raiz(1+x)

Substituindo x = 0

lim x-> Raiz(1+0)

= lim x-> 1 => 1

Resolvendo o lim

lim x -> 0 Raiz( 1 -x)/x

Usando regra de l'hospital.


lim x-> 0 F(x)'/G(x)'

F(x) = Raiz(1-x) = (1-x)^(1/2)

F(x)' = 1/2*(1-x)^(1/2-1)*(1-x)'

F(x)' = 1/2(1-x)^(-1/2)*-1

F(x)' = -1/2Raiz(1-x)
______

G(x) = x

G(x)' = 1

Então,

lim x ->0 -1/2Raiz(1-x)÷1

lim x -> 0 -1/2Raiz(1-0)

= -1/2
___________


Então teremos a subtração dos limites.

= 1 - (-1/2)

= 1+1/2

= 3/2
_____________


2)

lim x->1 3-Raiz(10-x)/(x^2-1)

Usando l'hospital:

lim x ->1 = F(x)'/G(x)'

F(x) = 3- Raiz(10-x)

F(x) = 3 -(10-x)^(1/2)

F(x)' = 0 -1/2(10-x)^(-1/2)(-1)

F(x)' = 1/2Raiz(10-x)
_______

G(x) = x^2 +1

G(x)' = 2x
_______

lim x-> 1 1/2Raiz(10-x)÷2

lim x->1 1/2Raiz(10-1)÷2

lim x-> 1 1/2Raiz(9)÷2

lim x->1 1/6÷2

lim x->1 1/12 = 1/12
__________


3)


Usando regra de l'hospital:

F(x) = x^3 -3x^2 +6x-4

F(x)' = 3x^2 -6x +6
_________

G(x) = x^3 -4x^2+8x-5

G(x)' = 3x^2 -8x +8
_________

Então,

lim x-> 1 F(x)'/G(x)'

lim x->1 (3x^2-6x+6)÷(3x^2-8x+8)

lim x->1 (3-6+6)/(3-8+8)

lim x-> 1 3/3 = 1