<s>[INST] Finding $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ this limit  according to wolframalpha is equal to $0$.
So this is my work thus far
$\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ output is $\infty - \infty$ which is indeterminate form.
So next I basically but it on the same denominator: $\frac{1}{3}$ $((3x + 2x^3 - 2(x^2+1)^{\frac{3}{2}})$ and turned $2(x^2+1)^{\frac{3}{2}}$ into something easier to work with $2\sqrt{x^2+1}+2x^{2}\sqrt{x^2+1}$
now the limit is $\frac{1}{3} \lim_{x \to \infty} ((3x + 2x^3-2\sqrt{x^2+1} -2x^{2}\sqrt{x^2+1})$ and this is where I am stuck to do next and lost.
 [/INST] First, observe
$$3x+2x^3-2(x^2+1)^{3/2}=\frac{3x^2+4}{-3x-2x^3-2\sqrt{x^2+1}-2x^2\sqrt{x^2+1}}.$$
The top is a quadratic, while the bottom grows on the order of $x^3$, hence the limit as $x\to \infty$ is zero.
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