Tính các giới hạn sau:

a) \(\mathop {\lim }\limits_{n \to  + \infty } \left( {\frac{{ - {n^2} + 7n - 9}}{{5{n^2} + 2n}}} \right)\); 

b) \[\mathop {\lim }\limits_{x \to 4} \frac{{\sqrt {x + 21}  - 5}}{{2x - 8}}\];

c) \(\mathop {\lim }\limits_{n \to  + \infty } \frac{{\sqrt {4{n^2} + n}  + 2n - 3}}{{n + 5}}\);   

d) \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2023x + 1} .\sqrt[3]{{2024x + 1}} - 1}}{x}\).

Lời giải

a) \(\mathop {\lim }\limits_{n \to  + \infty } \left( {\frac{{ - {n^2} + 7n - 9}}{{5{n^2} + 2n}}} \right)\)\[ = \mathop {\lim }\limits_{n \to  + \infty } \frac{{(\frac{{ - {n^2}}}{{{n^2}}} + \frac{{7n}}{{{n^2}}} - \frac{9}{{{n^2}}})}}{{(\frac{{5{n^2}}}{{{n^2}}} + \frac{{2n}}{{{n^2}}})}}\]\[ = \mathop {\lim }\limits_{n \to  + \infty } \frac{{( - 1 + \frac{7}{n} - \frac{9}{{{n^2}}})}}{{(5 + \frac{2}{n})}}\]\[ = \frac{{ - 1 + 0 - 0}}{{5 + 0}} = \frac{{ - 1}}{5}\].

b) \[\mathop {\lim }\limits_{x \to 4} \frac{{\sqrt {x + 21}  - 5}}{{2x - 8}}\]\[\mathop { = \lim }\limits_{x \to 4} \frac{{(\sqrt {x + 21}  - 5)(\sqrt {x + 21}  + 5)}}{{(2x - 8)(\sqrt {x + 21}  + 5)}} = \mathop {\lim }\limits_{x \to 4} \frac{{x + 21 - 25}}{{(2x - 8)(\sqrt {x + 21}  + 5)}}\]

\[ = \mathop {\lim }\limits_{x \to 4} \frac{{x - 4}}{{2(x - 4)(\sqrt {x + 21}  + 5)}} = \mathop {\lim }\limits_{x \to 4} \frac{1}{{2(\sqrt {x + 21}  + 5)}} = \frac{1}{{2(\sqrt {4 + 21}  + 5)}} = \frac{1}{{20}}\].

c)\[\mathop {\lim }\limits_{n \to  + \infty } \frac{{\sqrt {4{n^2} + n}  + 2n - 3}}{{n + 5}} = \mathop {\lim }\limits_{n \to  + \infty } \frac{{\sqrt {4 + \frac{1}{n}}  + 2 - \frac{3}{n}}}{{1 + \frac{5}{n}}}\]\[ = \frac{{\sqrt {4 + 0}  + 2 - 0}}{{1 + 0}} = 4.\]

d) \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2023x + 1} \sqrt[3]{{2024x + 1}} - 1}}{x}\)\( = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {2023x + 1} \sqrt[3]{{2024x + 1}} - \sqrt[3]{{2024x + 1}}} \right) + \left( {\sqrt[3]{{2024x + 1}} - 1} \right)}}{x}\)

\( = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2023x + 1} \sqrt[3]{{2024x + 1}} - \sqrt[3]{{2024x + 1}}}}{x} + \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{2024x + 1}} - 1}}{x}\)\[\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{2024x + 1}}\left( {\sqrt {2023x + 1}  - 1} \right)\left( {\sqrt {2023x + 1}  + 1} \right)}}{{x\left( {\sqrt {2023x + 1}  + 1} \right)}}\\ + \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[3]{{2024x + 1}} - 1} \right)\left( {\sqrt[3]{{{{(2024x + 1)}^2}}} + \sqrt[3]{{2024x + 1}} + 1} \right)}}{{x\left( {\sqrt[3]{{{{(2024x + 1)}^2}}} + \sqrt[3]{{2024x + 1}} + 1} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{2023.\sqrt[3]{{2024x + 1}}}}{{\left( {\sqrt {2023x + 1}  + 1} \right)}} + \mathop {\lim }\limits_{x \to 0} \frac{{2024}}{{\sqrt[3]{{{{(2024x + 1)}^2}}} + \sqrt[3]{{2024x + 1}} + 1}} = \frac{{2023}}{2} + \frac{{2024}}{3} = \frac{{10117}}{6}.\end{array}\]