Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}{x^2} - 1 + 2m\;khi\;\;x < 2\\\sqrt {x + 7} \;\;\;\;\;\;\;\;khi\;\;x \ge 2\end{array} \right.\) (m là tham số).

a) Khi m = −1 thì \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = 1\).

b) \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = 5\).

c) Tồn tại \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) khi m = −3.

d) \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = 3\).

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

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

c) \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{{x^3} - {x^2} + x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 2} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{x - 2}}{{{x^2} + 1}} = - \frac{1}{2} < 0\).

d) Để \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{ax + b}} = 2\) thì ax + b có nghiệm bằng 1 Û a + b = 0 Û b = −a.

Khi đó \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{ax + b}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 2} \right)\left( {x - 1} \right)}}{{a\left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{x - 2}}{a} = - \frac{1}{a} = 2\) \( \Leftrightarrow a = - \frac{1}{2} \Rightarrow b = \frac{1}{2}\).

Suy ra \(a + 3b = - \frac{1}{2} + 3.\frac{1}{2} = 1\).

Đáp án: a) Đúng; b) Sai;   c) Sai;   d) Đúng.

\(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 7ax + 5} + x} \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{{7ax + 5}}{{\sqrt {{x^2} + 7ax + 5} - x}}\)\( = \mathop {\lim }\limits_{x \to - \infty } \frac{{x\left( {7a + \frac{5}{x}} \right)}}{{x\left( { - \sqrt {1 + \frac{{7a}}{x} + \frac{5}{{{x^2}}}} - 1} \right)}} = - \frac{{7a}}{2}\).

Mà \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 7ax + 5} + x} \right) = - 3\) nên \( - \frac{{7a}}{2} = - 3\)\( \Leftrightarrow a = \frac{6}{7} \approx 0,86\).

Trả lời: 0,86.