Tìm giới hạn sau:a) \(\mathop {\lim }\limits_{x \to  + \infty } \left( { - {x^3} + 3{x^2} - 2x + 1} \right)\)      &nb

* Hướng dẫn giải

a) Ta có: \(\mathop {\lim }\limits_{x \to  + \infty } \left( { - {x^3} + 3{x^2} - 2x + 1} \right) = \mathop {\lim }\limits_{x \to  + \infty } {x^3}\left( { - 1 + \frac{3}{x} - \frac{2}{{{x^2}}} + \frac{1}{{{x^3}}}} \right)\)

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

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

b) 

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x + 1}  - 2}}{{9 - {x^2}}} = \mathop {\lim }\limits_{x \to 3} \frac{{(\sqrt {x + 1}  - 2)(\sqrt {x + 1}  + 2)}}{{(9 - {x^2})(\sqrt {x + 1}  + 2)}}\\
 = \mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}{{(9 - {x^2})(\sqrt {x + 1}  + 2)}} = \mathop {\lim }\limits_{x \to 3} \frac{{ - 1}}{{(3 + x)(\sqrt {x + 1}  + 2)}}\\
 = \frac{{ - 1}}{{(3 + 3)(\sqrt {3 + 1}  + 2)}} =  - \frac{1}{{24}}
\end{array}\)