Tính các giới hạn sau:a) A = \(\mathop {\lim }\limits_{x\, \to \,2} \;\frac{{4{x^2} + x - 18}}{{{x^3} - 8}}\)         

* Hướng dẫn giải

a) \(A = \mathop {\lim }\limits_{x\, \to \,2} \;\frac{{4{x^2} + x - 18}}{{{x^3} - 8}} = \mathop {\lim }\limits_{x\, \to \,2} \frac{{(x - 2)(4x + 9)}}{{(x - 2)({x^2} + 2x + 4)}} = \mathop {\lim }\limits_{x\, \to \,2} \frac{{4x + 9}}{{{x^2} + 2x + 4}} = \frac{{17}}{{12}}\)

b) 

\(\begin{array}{l}
B = \mathop {\lim }\limits_{x \to 2} \frac{{2 - \sqrt {x + 2} }}{{{x^2} - 3x + 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{(2 - \sqrt {x + 2} )(2 + \sqrt {x + 2} )}}{{({x^2} - 3x + 2)(2 + \sqrt {x + 2} )}} = \mathop {\lim }\limits_{x \to 2} \frac{{2 - x}}{{(x - 1)(x - 2)(2 + \sqrt {x + 2} )}}\\
        = \mathop {\lim }\limits_{x \to 2} \frac{{ - 1}}{{(x - 1)(2 + \sqrt {x + 2} )}} =  - \frac{1}{4}
\end{array}\)