<s>[INST] 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...
 [/INST] $$(\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} $$
Use Binomial theorem 
$$[x^3+5x^2]^{1/3} - [x^2 - 2x]^{1/2}$$
$$[x^3(1+\frac{5x^2}{x^3}]^{1/3}-[x^2(1-\frac{2x}{x^2}]^{1/2}$$
$$\approx x(1+\frac{5}{3x}.....)-x(1-\frac{1}{x}.....)$$
$$\approx [x+\frac{5}{3}]-[x-1]$$
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