# Difficulty to find this limit

1. Apr 22, 2012

### DDarthVader

Hello! I'm having difficulty to find this limit: $$\lim_{x\rightarrow -1}(\sqrt[3]{\frac{x^3+1}{x+1}})$$
This is what I'm trying to do to solve this limit: Let $$\frac{x^3+1}{x+1}=u$$ then
$$\lim_{x\rightarrow -1}(\sqrt[3]{u}) = \lim_{x\rightarrow -1}(\sqrt[3]{-1})$$

I know something is wrong I'm just not sure about what is actually wrong. I'm thinking that $$\lim_{x\rightarrow -1}(\sqrt[3]{u})$$ Is wrong because that -1!

Thanks!

2. Apr 22, 2012

### DonAntonio

Hint: $x^3+1=(x+1)(x^2-x+1)$ .

DonAntonio

3. Apr 22, 2012

### Useful nucleus

Also when you use the algebraic substitution, you need to consider the limit of the new variable u. So, as x--> -1 , what is the limiting value of u? However, this line of thiking will not save you from facing the indeterminate form (0/0). The easiest path is the hint that DonAntonio suggested.

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