<s>[INST] Finding $\lim\limits _{x\to-\infty} \frac{\sqrt[3]{x^3+x}}{x}$ 
$$\lim\limits _{x\to-\infty} \frac{\sqrt[3]{x^3+x}}{x}$$

I have to resolve this limit, I tried factoring out x, I tried rewriting $x^3+x$ as 
$x^3(1+ \frac{x}{x^3})$ and it doesn't seem to cancel. What should I do?
 [/INST] \begin{align}
\lim_{x\to -\infty} \frac{\sqrt[\,3]{x^3+x}}{x}
=&
\lim_{x\to -\infty} \frac{\sqrt[\,3]{x^3\cdot(1+1/x^2)}}{x}
\\
=&
\lim_{x\to -\infty} \frac{\sqrt[\,3]{x^3}\cdot\sqrt[\,3]{(1+1/x^2)}}{x}
\\
=&
\lim_{x\to -\infty} \frac{x\cdot\sqrt[\,3]{(1+1/x^2)}}{x}
\\
=&
\lim_{x\to -\infty} {\sqrt[\,3]{(1+1/x^2)}}
\\
=&
{\sqrt[\,3]{\lim_{x\to -\infty} (1+1/x^2)}}
\\
=&
 {\sqrt[\,3]{(1+\lim_{x\to -\infty} (1/x^2))}}
\\
=&
 {\sqrt[\,3]{(1+0)}}
\\
=&
 {\sqrt[\,3]{1}}
\\
=&
1
\end{align}
</s>