Tính các giới hạn sau:a) \(\lim \frac{{4n + 9}}{{6n - 7}}\)b) \(\mathop {\lim }\limits_{x \to \, - 2} \frac{{2 - 5x - 3{x^2}}}{{2x + 4}}\)

* Hướng dẫn giải

a) \(\lim \frac{{4n + 9}}{{6n - 7}} = \lim \frac{{n\left( {4 + \frac{9}{n}} \right)}}{{n\left( {6 - \frac{7}{n}} \right)}} = \lim \frac{{4 + \frac{9}{n}}}{{6 - \frac{7}{n}}} = \frac{4}{6} = \frac{2}{3}\)

b) \(\mathop {\lim }\limits_{x \to  - 2} \frac{{2 - 5x - 3{x^2}}}{{2x + 4}} = \mathop {\lim }\limits_{x \to  - 2} \frac{{3\left( {x - \frac{1}{3}} \right)\left( {x + 2} \right)}}{{2\left( {x + 2} \right)}} = \mathop {\lim }\limits_{x \to  - 2} \frac{{3\left( {x - \frac{1}{3}} \right)}}{2} = \frac{{3\left( { - 2 - \frac{1}{3}} \right)}}{2} =  - \frac{7}{2}\)