<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] When $x$ is negative, then $\sqrt{x^2} = -x$, which is positive.  So we have
\begin{align}
& \lim_{x\to-\infty} \left(\sqrt{x^2+2x}+\sqrt[3]{x^3+x^2}\right) \\[10pt]
= {} & \lim_{x\to-\infty} \left( -x\sqrt{1 + \frac 2 x} + x \sqrt[3]{1 +  \frac 1 x} \, \right) \\[10pt]
= {} & \lim_{x\to-\infty} x\left( -\left( 1 + \frac 2 x \right)^{1/2} + \left( 1 + \frac 1 x \right)^{1/3} \right) \\[10pt]
 = {} & \lim_{x\to-\infty} x\left( -\left( 1 + \frac 2 x \right)^{1/2} + \left( 1 + \frac 1 x \right)^{1/3} \right) \\[10pt]
 = {} & \lim_{u\,\uparrow\, 0} \frac 1 u \left( - (1+2u)^{1/2} + (1+u)^{1/3} \right)  \text{ where } u = \frac 1 x.
\end{align}
L'Hopital's rule can deal with that. (And remember the chain rule when finding $\dfrac d {du} (1+2u)^{1/2}.$)
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