<s>[INST] The limit of a function with sum of two roots. I need to find the limit of the following function:
$$\lim_{x\to-\infty} \left(\sqrt{x^2+2x}+\sqrt[3]{x^3+x^2}\right).$$
I derived it to the form of:
$$\lim_{x\to-\infty}\frac{(x^2+2x)^3-(x^3+x^2)^2}{\left(\sqrt{x^2+2x}-\sqrt[3]{x^3+x^2}\right)\left((x^2+2x)^2+(x^2+2x)\sqrt[3]{(x^3+x^2)^2}+\sqrt[3]{(x^3+x^2)^4}\right)},$$
hoping that I'd be able to simplify something but still have $\frac{0}{0}$ and don't see how to do that.
 [/INST] Let me change the sign on $x$, I find it easier to think about.
the expression is then
$$\sqrt{x^2-2x}+\sqrt[3]{x^2-x^3}$$
Now investigate the two limits 
$$\sqrt{x^2-2x}-x\rightarrow -1$$
and 
$$x+\sqrt[3]{x^2-x^3}$$ and this latter is equal
to 
$$\frac{x^2}{x^2-x\sqrt[3]{x^2-x^3}+(\sqrt[3]{x^2-x^3})^2}$$
for which I get the limit  of $\frac{1}{3}$ (assuming I haven't made a mistake, (a big assumption)).
Ok so I get the limit $-\frac{2}{3}$ in agreement with the book. 
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