Find $\mathop {\lim }\limits_{x \to \infty } (\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} ) $ $$\mathop {\lim }\limits_{x \to \infty } (\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} )
$$
My try:
$${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})
$$
$$\mathop {\lim }\limits_{x \to \infty } \frac{{(\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} )(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x}  + {x^2} - 2x)}}{{(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x}  + {x^2} - 2x)}} = 
$$
$$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3} + 5{x^2} - {x^2} - 2x}}{{(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x}  + {x^2} - 2x)}}
$$
And what's next...?
This task in first and second remarkable limits. I think i can replace variable, but how i will calculate it...
 A: If you can use the Taylor expansion $(1+t)^\alpha=1+\alpha t + \hbox{higher order terms}$ as $t\to0$, then it is easy:
$$
\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} =x\left(\sqrt[3]{{1 + 5/x}} - \sqrt {1 - 2/x}\right)=x\left(1 + 5/(3x) - 1 + 1/x+ \hbox{h. o. t.}\right)=8/3+ \hbox{h. o. t.}\to8/3
$$
A: We can also solve Using $$\displaystyle \bullet\; (1+x)^{n} = 1+nx+\frac{n(n-1)x^2}{2}+..........+\infty$$
So Here Limit $$\displaystyle \lim_{x\rightarrow \infty}\left[\sqrt[3]{x^3+5x^2}-\sqrt{x^2-2x}\right] = \lim_{x\rightarrow \infty}x\cdot \left[\left(1+\frac{5}{x}\right)^{\frac{1}{3}}-\left(1-\frac{2}{x}\right)^{\frac{1}{2}}\right]$$
So limit $$\displaystyle \lim_{x\rightarrow \infty}x\cdot \left[\left(1+\frac{5}{3x}+\frac{1}{3}\cdot -\frac{2}{3}\cdot \frac{25}{x^2}+........\infty\right)-\left(1-\frac{2}{2x}+\frac{1}{2}\cdot -\frac{1}{2}\cdot \frac{4}{x^2}.......\infty\right)\right]$$
So we get $$\displaystyle \lim_{x\rightarrow \infty}x\cdot \left[\frac{5}{3x}+\frac{1}{x}\right] = \frac{8}{3}$$
Here we Negelect Higher power of $x$ in Denominator, bcz when $\lim_{x\rightarrow \infty}\;,$ Then $\displaystyle \frac{1}{x^n}\rightarrow 0\; \forall n\geq 1$
A: The conjugate expression of a mix of cube and square roots is quite complicated. It is much simpler to use Taylor's formula at order $1$:
\begin{align*}
\sqrt[3]{x^3+5x}&=x\sqrt[3]{1+\frac5x}=x\biggl(1+\frac5{3x} +o\Bigl(\frac1x\Bigr)\biggr)=x+\frac53+o(1)\\
\sqrt{x^2-2x}&=x\sqrt{1-\frac2x}=x\biggl(1-\frac1x +o\Bigl(\frac1x\Bigr)\biggr)=x-1+o(1)
\end{align*}
(we may suppose $x>0$), hence:
$$\sqrt[3]{x^3+5x}-\sqrt{x^2-2x}=\frac83+o(1)\xrightarrow[x\to+\infty]{}\frac83.$$
A: $$(\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} $$
Use Binomial theorem 
$$[x^3+5x^2]^{1/3} - [x^2 - 2x]^{1/2}$$
$$[x^3(1+\frac{5x^2}{x^3}]^{1/3}-[x^2(1-\frac{2x}{x^2}]^{1/2}$$
$$\approx x(1+\frac{5}{3x}.....)-x(1-\frac{1}{x}.....)$$
$$\approx [x+\frac{5}{3}]-[x-1]$$
