Tính các giới hạn sau:

a) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{5n + 4}}{{20n - 3}}\)                                 b) \(\mathop {\lim }\limits_{x \to - \infty } \left( { - {x^3} + 3{x^2} - 6} \right)\)

c) \[\mathop {\lim }\limits_{x \to - 3} \frac{{\sqrt {{x^2} + 7} - 4}}{{x + 3}}\]                              d) \[A = \mathop {\lim }\limits_{x \to 0} {\kern 1pt} \frac{{\sqrt[3]{{x + 1}}\sqrt {x + 4} - 2}}{x}\]

a)\(\mathop {\lim }\limits_{n \to + \infty } \frac{{5n + 4}}{{20n - 3}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{5 + \frac{4}{n}}}{{20 - \frac{3}{n}}} = \frac{{\mathop {\lim }\limits_{n \to + \infty } \left( {5 + \frac{4}{n}} \right)}}{{\mathop {\lim \left( {20 - \frac{3}{n}} \right)}\limits_{} }} = \frac{1}{5}\)

b)\(\mathop {\lim }\limits_{x \to - \infty } \left( { - {x^3} + 3{x^2} - 6} \right) = \mathop {\lim }\limits_{x \to - \infty } {x^3}\left( { - 1 + \frac{3}{x} - \frac{6}{{{x^2}}}} \right) = + \infty \)

Vì \(\mathop {\lim }\limits_{x \to - \infty } {x^3} = - \infty ,\,\,\mathop {\lim }\limits_{x \to - \infty } \left( { - 1 + \frac{3}{x} - \frac{6}{{{x^2}}}} \right) = - 1 < 0\)

c)\[\begin{array}{l}\mathop {\lim }\limits_{x \to - 3} \frac{{\sqrt {{x^2} + 7} - 4}}{{x + 3}} = \mathop {\lim }\limits_{x \to - 3} \frac{{{x^2} + 7 - 16}}{{\left( {x + 3} \right)\left( {\sqrt {{x^2} + 7} + 4} \right)}} = \mathop {\lim }\limits_{x \to - 3} \frac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {\sqrt {{x^2} + 7} + 4} \right)}}\\ = \mathop {\lim }\limits_{x \to - 3} \frac{{x - 3}}{{\left( {\sqrt {{x^2} + 7} + 4} \right)}} = \frac{{ - 6}}{8} = - \frac{3}{4}\end{array}\]

d)\[A = \mathop {\lim }\limits_{x \to 0} {\kern 1pt} \frac{{\sqrt[3]{{x + 1}}\sqrt {x + 4} - 2}}{x} = \mathop {\lim }\limits_{x \to 0} {\kern 1pt} \frac{{\sqrt[3]{{x + 1}}\sqrt {x + 4} - \sqrt {x + 4} + \sqrt {x + 4} - 2}}{x}\]

\[ = \mathop {\lim }\limits_{x \to 0} {\kern 1pt} \frac{{\sqrt {x + 4} \left( {\sqrt[3]{{x + 1}} - 1} \right) + \sqrt {x + 4} - 2}}{x} = \mathop {\lim }\limits_{x \to 0} {\kern 1pt} \left( {\frac{{\sqrt {x + 4} \left( {\sqrt[3]{{x + 1}} - 1} \right)}}{x} + \frac{{\sqrt {x + 4} - 2}}{x}} \right)\]

Xét:

\[B = \mathop {\lim }\limits_{x \to 0} {\kern 1pt} \frac{{\sqrt {x + 4} \left( {\sqrt[3]{{x + 1}} - 1} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} {\kern 1pt} \frac{{\sqrt {x + 4} \left( {x + 1 - 1} \right)}}{{x\left( {{{\left( {\sqrt[3]{{x + 1}}} \right)}^2} + \sqrt[3]{{x + 1}} + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} {\kern 1pt} \frac{{x\sqrt {x + 4} }}{{x\left( {{{\left( {\sqrt[3]{{x + 1}}} \right)}^2} + \sqrt[3]{{x + 1}} + 1} \right)}}\]

\[\mathop {\lim }\limits_{x \to 0} {\kern 1pt} \frac{{\sqrt {x + 4} }}{{{{\left( {\sqrt[3]{{x + 1}}} \right)}^2} + \sqrt[3]{{x + 1}} + 1}} = \frac{2}{3}\]

\[C = \mathop {\lim }\limits_{x \to 0} {\kern 1pt} \frac{{\sqrt {x + 4} - 2}}{x} = \mathop {\lim }\limits_{x \to 0} {\kern 1pt} \frac{x}{{x\left( {\sqrt {x + 4} + 2} \right)}} = \mathop {\lim }\limits_{x \to 0} {\kern 1pt} \frac{1}{{\sqrt {x + 4} + 2}} = \frac{1}{4}\]

Vậy \[A = B + C = \frac{2}{3} + \frac{1}{4} = \frac{{11}}{{12}}\]