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Answer 1 · 2016-11-10T12:17:35

$1/3$

Making $x = y^3$

$(x^(1/3) +1)/(x+1) equiv (y+1)/(y^3+1)$ but by the polynomial identity

$(a^3+1)/(a+1)=a^2-a+1$ we have

$(x^(1/3) +1)/(x+1) equiv 1/(y^2-y+1)$

If we are interested only in real solutions, then

$lim_(x->-1)(x^(1/3) +1)/(x+1) equiv lim_(y->-1)1/(y^2-y+1) = 1/3$

Answer 2 · 2016-11-10T14:12:54

1/3

The indeterminate form is 0/0.

L'Hospital's rule can be applied.

The limit is

$lim x to -1 ((x^(1/3)+1)^')/((x+1)')=lim x to -1 ((1/3)(x^(-2))^(1/3))/1=((-1)^2)^(1/3)=1/3$