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...

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:
$$\sqrt[3]{x^3+5x}-\sqrt{x^2-2x}=\frac83+o(1)\xrightarrow[x\to+\infty]{}\frac83.$$
