a) Tính \(\mathop {\lim }\limits_{x \to + \infty } \frac{{2{x^2} - 3x + 1}}{{{x^2} - 1}}\);

b) Cho hàm số \(f\left( x \right)\) xác định trên \(\mathbb{R}\) thỏa mãn \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 5}}{{x - 2}} = 3.\) Tìm \(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {f\left( x \right) + 4} - 3}}{{{x^2} + x - 6}}\)

a) Ta có \[\mathop {\lim }\limits_{x \to + \infty } \frac{{2{x^2} - 3x + 1}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{2 - \frac{3}{x} + \frac{1}{{{x^2}}}}}{{1 - \frac{1}{{{x^2}}}}}\]\[ = \frac{{\mathop {\lim }\limits_{x \to + \infty } 2 - \mathop {\lim }\limits_{x \to + \infty } \frac{3}{x} + \mathop {\lim }\limits_{x \to + \infty } \frac{1}{{{x^2}}}}}{{\mathop {\lim }\limits_{x \to + \infty } 1 - \mathop {\lim }\limits_{x \to + \infty } \frac{1}{{{x^2}}}}} = \frac{{2 - 0 + 0}}{{1 - 0}} = 2\]

b) Nếu \(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) - 5} \right] \ne 0\) thì \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 5}}{{x - 2}} = + \infty \) hoặc\(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 5}}{{x - 2}} = - \infty \). Mâu thuẫn giả thiết . Vậy \(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) - 5} \right] = 0\).Suy ra \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 5.\)

Ta có :

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {f\left( x \right) + 4} - 3}}{{{x^2} + x - 6}} = \mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) + 4 - 9}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left( {\sqrt {f\left( x \right) + 4} + 3} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 5}}{{\left( {x - 2} \right)}}.\mathop {\lim }\limits_{x \to 2} \frac{1}{{\left( {x + 3} \right)\left( {\sqrt {f\left( x \right) + 4} + 3} \right)}}\\ = 3.\frac{1}{{5.6}} = \frac{1}{{10}}\end{array}\)