Đáp án:
$\begin{array}{l}
v)\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 1} - \sqrt[3]{{{x^3} - 1}}} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 1} - x + x - \sqrt[3]{{{x^3} - 1}}} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{{x^2} + 1 - {x^2}}}{{\sqrt {{x^2} + 1} + x}} + \frac{{{x^3} - {x^3} + 1}}{{{x^2} + x\sqrt[3]{{{x^3} - 1}} + \sqrt[3]{{{{\left( {{x^3} + 1} \right)}^2}}}}}} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {\frac{1}{{\sqrt {{x^2} + 1} + x}} + \frac{1}{{{x^2} + x\sqrt[3]{{{x^3} - 1}} + \sqrt[3]{{{{\left( {{x^3} + 1} \right)}^2}}}}}} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{\frac{1}{x}}}{{\sqrt {1 + \frac{1}{{{x^2}}}} + 1}} + \frac{{\frac{1}{{{x^2}}}}}{{1 + \sqrt[3]{{1 - \frac{1}{{{x^3}}}}} + {{\left( {\sqrt[3]{{1 - \frac{1}{{{x^3}}}}}} \right)}^2}}}} \right)\\
= 0\\
{\rm{w}})\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt[3]{{{x^3} + 2x - 1}} - \sqrt {{x^2} - 3x} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt[3]{{{x^3} + 2x - 1}} - x + x - \sqrt {{x^2} - 3x} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{{x^3} + 2x - 1 - {x^3}}}{{{{\left( {\sqrt[3]{{{x^3} + 2x - 1}}} \right)}^2} + x\sqrt[3]{{{x^3} + 2x - 1}} + {x^2}}} + \frac{{{x^2} - {x^2} + 3x}}{{x + \sqrt {{x^2} - 3x} }}} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{2x - 1}}{{{{\left( {\sqrt[3]{{{x^3} + 2x - 1}}} \right)}^2} + x\sqrt[3]{{{x^3} + 2x - 1}} + {x^2}}} + \frac{{3x}}{{x + \sqrt {{x^2} - 3x} }}} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{\frac{2}{x} - \frac{1}{{{x^2}}}}}{{{{\left( {\sqrt[3]{{1 + \frac{2}{{{x^2}}} - \frac{1}{{{x^3}}}}}} \right)}^2} + \sqrt[3]{{1 + \frac{2}{{{x^2}}} - \frac{1}{{{x^3}}}}} + 1}} + \frac{3}{{1 + \sqrt {1 - \frac{3}{x}} }}} \right)\\
= \frac{3}{2}
\end{array}$
