Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
i,\\
\mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x - 2}}{{\sqrt {x + 5} - \sqrt {2x + 1} }}\\
= \mathop {\lim }\limits_{x \to 4} \frac{{\frac{{x - 4}}{{\sqrt x + 2}}}}{{\frac{{\left( {x + 5} \right) - \left( {2x + 1} \right)}}{{\sqrt {x + 5} + \sqrt {2x + 1} }}}}\\
= \mathop {\lim }\limits_{x \to 4} \left( {\frac{{x - 4}}{{\sqrt x + 2}}:\frac{{ - \left( {x - 4} \right)}}{{\sqrt {x + 5} + \sqrt {2x + 1} }}} \right)\\
= - \mathop {\lim }\limits_{x \to 4} \frac{{\sqrt {x + 5} + \sqrt {2x + 1} }}{{\sqrt x + 2}}\\
= - \frac{{\sqrt {4 + 5} + \sqrt {2.4 + 1} }}{{\sqrt 4 + 2}}\\
= \frac{{ - 6}}{4} = - \frac{3}{2}\\
k,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 2x} - \sqrt[3]{{{x^3} + 3{x^2}}}} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {\left( {\sqrt {{x^2} + 2x} - x} \right) + \left( {x - \sqrt[3]{{{x^3} + 3{x^2}}}} \right)} \right]\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {\frac{{{x^2} + 2x - {x^2}}}{{\sqrt {{x^2} + 2x} + x}} + \frac{{{x^3} - {x^3} - 3{x^2}}}{{{x^2} + x.\sqrt[3]{{{x^3} + 3{x^2}}} + {{\sqrt[3]{{{x^3} + 3{x^2}}}}^2}}}} \right]\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {\frac{{2x}}{{\sqrt {{x^2} + 2x} + x}} + \frac{{ - 3{x^2}}}{{{x^2} + x.\sqrt[3]{{{x^3} + 3{x^2}}} + {{\sqrt[3]{{{x^3} + 3{x^2}}}}^2}}}} \right]\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {\frac{2}{{\sqrt {1 + \frac{2}{x}} + 1}} + \frac{{ - 3}}{{1 + 1.\sqrt[3]{{1 + \frac{3}{x}}} + {{\sqrt[3]{{1 + \frac{3}{x}}}}^2}}}} \right]\\
= \frac{2}{{\sqrt 1 + 1}} + \frac{{ - 3}}{{1 + 1 + 1}}\\
= 1 - 1 = 0
\end{array}\)
