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.

Hint:
WLOG $x=\tan y\implies y\to\dfrac\pi2$
$$\dfrac{3\tan y+2\tan^3y-2\sec^3y}3$$
$$=\dfrac{3\sin y\cos^2y+2\sin^3y-2}{3\cos^3y}$$
The numerator $$=3(1-\sin^2y)\sin y+2\sin^3y-2=\cdots=(1-\sin y)^2(2\sin y+1)$$
Finally use $$\dfrac{1-\sin y}{\cos y}=\dfrac{\cos y}{1+\sin y}$$
