Tìm các giới hạn sau:1) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 3x - 2}}{{{x^2} - 4}}\)2) \(\mathop {\lim }\limits_{x \t

* Hướng dẫn giải

1) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 3x - 2}}{{{x^2} - 4}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {{x^2} + 2x + 1} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 2x + 1}}{{x + 2}} = \frac{9}{4}\)

2) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{1 + 2x - {x^3}}}{{{x^3} - 3{x^2} + 5}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\frac{1}{{{x^3}}} + \frac{2}{{{x^2}}} - 1}}{{1 - \frac{3}{x} + \frac{5}{{{x^3}}}}} =  - 1\)

3) \(\mathop {\lim }\limits_{x \to {3^ - }} \frac{{2x + 3}}{{x - 3}} =  - \infty \)

Vì \(\left\{ \begin{array}{l}
\mathop {\lim }\limits_{x \to {3^ - }} \left( {2x + 3} \right) = 9 > 0\\
\mathop {\lim }\limits_{x \to {3^ - }} \left( {x - 3} \right) = 0\\
x - 3 < 0,\forall x < 3
\end{array} \right.\)