Question

lim √x+2 ao cubo -1/x+1 , x tende a (-1)​

Answer 1

lim  [√(x+2)³ -1]/(x+1)

x-->-1

(a-b)³=a³-3a²b+3ab²-b³

(a-b)³=a³-b³-3ab(a-b)

a³-b³=(a-b)*[(a-b)²+3ab]

a³-b³=(a-b)*(a²+ab+b³)

Fazendo a=√(x+2)  e b=1

√(x+2)³ -1 =(√(x+2)-1)*(√(x+2)²+√(x+2)+1)

lim  [(√(x+2)-1)*(√(x+2)²+√(x+2)+1)]/(x+1)

x-->-1

lim  [(√(x+2)-1)*(√(x+2)²+√(x+2)+1)(√(x+2)+1)]/[(x+1)(√(x+2)+1)]

x-->-1

lim  [(√(x+2)²-1²)*(√(x+2)²+√(x+2)+1)]/[(x+1)(√(x+2)+1)]

x-->-1

lim  [(x+2-1)*(√(x+2)²+√(x+2)+1)]/[(x+1)(√(x+2)+1)]

x-->-1

lim  [(x+1)*(√(x+2)²+√(x+2)+1)]/[(x+1)(√(x+2)+1)]

x-->-1

lim  [(√(x+2)²+√(x+2)+1)]/(√(x+2)+1)

x-->-1

=(1+1+1)/(1+1)

=3/2 é a resposta