Giải thích các bước giải:
\(\begin{array}{l}
3,\\
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {3x + 4} - \sqrt[3]{{8 + 5x}}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {3x + 4} - 2} \right) + \left( {2 - \sqrt[3]{{8 + 5x}}} \right)}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{3x + 4 - {2^2}}}{{\sqrt {3x + 4} + 2}} + \frac{{{2^3} - \left( {8 + 5x} \right)}}{{4 + 2\sqrt[3]{{8 + 5x}} + {{\sqrt[3]{{8 + 5x}}}^2}}}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{3x}}{{\sqrt {3x + 4} + 2}} - \frac{{5x}}{{4 + 2\sqrt[3]{{8 + 5x}} + {{\sqrt[3]{{8 + 5x}}}^2}}}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \left[ {\frac{3}{{\sqrt {3x + 4} + 2}} - \frac{5}{{4 + 2\sqrt[3]{{8 + 5x}} + {{\sqrt[3]{{8 + 5x}}}^2}}}} \right]\\
= \frac{3}{{\sqrt {3.0 + 4} + 2}} - \frac{5}{{4 + 2\sqrt[3]{{8 + 5.0}} + {{\sqrt[3]{{8 + 5.0}}}^2}}}\\
= \frac{3}{4} - \frac{5}{{12}}\\
= \frac{1}{3}\\
4,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt[3]{{{x^3} + 2{x^2}}} - \sqrt {{x^2} - 2x} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {\left( {\sqrt[3]{{{x^3} + 2{x^2}}} - x} \right) + \left( {x - \sqrt {{x^2} - 2x} } \right)} \right]\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {\frac{{{x^3} + 2{x^2} - {x^3}}}{{{{\sqrt[3]{{{x^3} + 2{x^2}}}}^2} + x.\sqrt[3]{{{x^3} + 2{x^2}}} + {x^2}}} + \frac{{{x^2} - \left( {{x^2} - 2x} \right)}}{{x + \sqrt {{x^2} - 2x} }}} \right]\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {\frac{{2{x^2}}}{{{{\sqrt[3]{{{x^3} + 2{x^2}}}}^2} + x.\sqrt[3]{{{x^3} + 2{x^2}}} + {x^2}}} + \frac{{2x}}{{x + \sqrt {{x^2} - 2x} }}} \right]\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {\frac{2}{{{{\sqrt[3]{{1 + \frac{2}{x}}}}^2} + 1.\sqrt[3]{{1 + \frac{2}{x}}} + {1^2}}} + \frac{2}{{1 + \sqrt {1 - \frac{2}{x}} }}} \right]\\
= \frac{2}{{{{\sqrt[3]{1}}^2} + 1.1 + {1^2}}} + \frac{2}{{1 + \sqrt 1 }}\\
= \frac{2}{3} + 1 = \frac{5}{3}
\end{array}\)
