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.
 A: $$x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3}=\frac{\left(x+\frac{2}{3}x^3\right)^2-\frac{4}{9}(x^2+1)^3}{x + \frac{2x^{3}}{3}+ \frac{2(x^2+1)^{\frac{3}{2}}}{3}}=$$
$$=\frac{-\frac{1}{3}x^2-\frac{4}{9}}{x + \frac{2x^{3}}{3}+ \frac{2(x^2+1)^{\frac{3}{2}}}{3}}=\frac{-\frac{1}{3x}-\frac{4}{9x^3}}{\frac{1}{x^2} + \frac{2}{3}+ \frac{2(1+\frac{1}{x^2})^{\frac{3}{2}}}{3}}\rightarrow0$$ for $x\rightarrow+\infty.$
A: 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.
A: By binomial approximation
$$(x^2+1)^{\frac{3}{2}}=(x^2)^{\frac{3}{2}}\left(1+\frac1{x^2}\right)^{\frac{3}{2}} = x^3+\frac32 x +O\left(\frac1{x}\right)\implies \frac{2(x^2+1)^{\frac{3}{2}}}{3} = \frac{2x^{3}}{3}+x+O\left(\frac1{x}\right)$$
therefore
$$x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3}=O\left(\frac1{x}\right)\to 0$$
A: 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 $$\dfrac{1-\sin y}{\cos y}=\dfrac{\cos y}{1+\sin y}$$
A: You should recall that $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$, since $(c-d)(c+d)=c^2-d^2$. This should help you simplify the expression with the square root.
Though I don't understand how you obtained what you wrote, from what I can see you should get:
$$ x+ \frac{2x^3}{3}-\frac{2(x^2+1)^{\frac{3}{2}}}{3}=x +\frac{2}{3}\Big( \frac{x^6-(x^2+1)^3}{x^3 +(x^2+1)^{ \frac{3}{2} } } \Big)=x+\frac{2}{3}\frac{-3x^4-3x^2-1}{x^3+(x^2+1)^{ \frac{3}{2} }}$$
A: $$A=x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3}=x + \frac{2x^{3}}{3} - \frac{2x^3}{3}\left(1+\frac{1}{x^2}\right)^{3/2}$$ For the very last term, let $\frac{1}{x^2}=\epsilon$ and use the binomial expansion
$$(1+\epsilon)^{3/2}=1+\frac{3 \epsilon }{2}+\frac{3 \epsilon ^2}{8}+O\left(\epsilon ^3\right)$$ Replace $\epsilon$ by $\frac{1}{x^2}$ to make
$$\left(1+\frac{1}{x^2}\right)^{3/2}=1+\frac{3}{2 x^2}+\frac{3}{8 x^4}+O\left(\frac{1}{x^6}\right)$$
$$A=x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3}=x + \frac{2x^{3}}{3} - \frac{2x^3}{3}\left(1+\frac{3}{2 x^2}+\frac{3}{8 x^4}+O\left(\frac{1}{x^6}\right)\right)$$
$$A=-\frac{1}{4 x}+O\left(\frac{1}{x^3}\right)$$
