Question

lim (raiz cubica de x-1)/raiz quarta de x-1
x->1

como resolvo esse limite? 

Answer 1

(a-1)³=a³-3a²+3a-1
(a-1)³=a³-1-3a(a-1)
a³-1=(a-1)³-3a(a-1)
a³-1=(a-1)*[(a-1)²+3a]
a³-1=(a-1)*[a²-2a+1+3a]
a³-1=(a-1)*(a²+1+a)

Fazendo a=∛x, temos então:

∛x³-1=(∛x-1)*[∛x²+1+∛x]
x-1=(∛x-1)*[∛x²+1+∛x]
(∛x-1) = (x-1)/[∛x²+1+∛x]

####################
x-1=(√×+1)(√x-1)
x-1=(√×+1) (x^(1/4) +1)(x^(1/4) -1)

(x^(1/4) -1) =(x-1)/[(√×+1) (x^(1/4) +1)]


Lim (∛x-1)/(x^(1/4)-1)
x->1

Lim {(x-1)/(∛x²+1+∛x)}/{(x-1)/[(√×+1) (x^(1/4) +1)]}
x->1

Lim {(x-1)/(∛x²+1+∛x)}*{[(√×+1) (x^(1/4) +1)/(x-1)]}
x->1

Lim {1/(∛x²+1+∛x)}*{[(√×+1) (x^(1/4) +1)}
x->1

= {1/(∛1²+1+∛1)}*{[(√1+1) (1^(1/4) +1)}


= {1/(1+1+1)}*{[(1+1) * (1 +1)}

= {1/(3)}*{[(2)* (2)}   =4/3 é a resposta