Question

$\lim_{x \to 0} \frac{\sqrt[3]{8+x}-2}{x}$ sem usar L'Hopital

Answer 1

(a-b)³=a³-3a²b+3ab²-b³

(a-b)³=a³-b³-3ab*(a-b)

a³-b³ =(a-b)³+3ab*(a-b)

a³-b³ = (a-b)*[(a-b)²+3ab]

a³-b³ =(a-b) * (a²-2ab+b²+3ab)

a³-b³=(a-b)*(a²+ab+b²)

(a-b) =(a³-b³)/(a²+ab+b²)

Fazendo a=∛(8+x)    e b=2

∛(8+x)  - 2 = (8+x-8)/(∛(8+x)²+2*∛(8+x) +2²)

∛(8+x)  - 2 = x /(∛(8+x)²+2*∛(8+x) +4)

Lim   [∛(8+x) -2]/x

x-->0

Lim   x /x(∛(8+x)²+2*∛(8+x) +4)

x-->0

Lim   1 /(∛(8+x)²+2*∛(8+x) +4)

x-->0

=1 /(∛(8+0)²+2*∛(8+0) +4)

=1 /(∛8²+2*∛2³ +4)

=1/(4+4+4) =1/12