# Question #7b587

Nov 10, 2016

$\frac{1}{3}$

#### Explanation:

Making $x = {y}^{3}$

$\frac{{x}^{\frac{1}{3}} + 1}{x + 1} \equiv \frac{y + 1}{{y}^{3} + 1}$ but by the polynomial identity

$\frac{{a}^{3} + 1}{a + 1} = {a}^{2} - a + 1$ we have

$\frac{{x}^{\frac{1}{3}} + 1}{x + 1} \equiv \frac{1}{{y}^{2} - y + 1}$

If we are interested only in real solutions, then

${\lim}_{x \to - 1} \frac{{x}^{\frac{1}{3}} + 1}{x + 1} \equiv {\lim}_{y \to - 1} \frac{1}{{y}^{2} - y + 1} = \frac{1}{3}$

Nov 10, 2016

1/3

#### Explanation:

The indeterminate form is 0/0.

L'Hospital's rule can be applied.

The limit is

$\lim x \to - 1 \frac{{\left({x}^{\frac{1}{3}} + 1\right)}^{'}}{\left(x + 1\right) '} = \lim x \to - 1 \frac{\left(\frac{1}{3}\right) {\left({x}^{- 2}\right)}^{\frac{1}{3}}}{1} = {\left({\left(- 1\right)}^{2}\right)}^{\frac{1}{3}} = \frac{1}{3}$