Cho hàm số f(x) xác định trên ℝ và thỏa mãn \(\mathop {\lim }\limits_{x \to - 1} \frac{{f\left( x \right) - 2}}{{x + 1}} = 2024\). Giới hạn \[\mathop {\lim }\limits_{x \to - 1} \frac{{{f^2}\left( x \right) + f\left( x \right) - 6}}{{x + 1}}\] bằng     

a) \(\mathop {\lim }\limits_{x \to 2} 3f\left( x \right) = 3\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 3.2024 = 6072\).

b) \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right)}}{4} = \frac{{2024}}{4} = 506\).

c) \(\mathop {\lim }\limits_{x \to 2} \sqrt {f\left( x \right)}  = \sqrt {2024}  = 2\sqrt {506} \).

d) \(\mathop {\lim }\limits_{x \to 2} \left[ {100x - \frac{1}{2}f\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 2} 100x - \frac{1}{2}\mathop {\lim }\limits_{x \to 2} f\left( x \right)\)\( = 200 - \frac{1}{2}.2024 =  - 812\).

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

\(\mathop {\lim }\limits_{x \to  + \infty } \left( {\frac{{{x^2} + 3x + 1}}{{x + 1}} + ax + b} \right) = 1\)\( \Leftrightarrow \mathop {\lim }\limits_{x \to  + \infty } \left( {\frac{{\left( {a + 1} \right){x^2} + \left( {a + b + 3} \right)x + b + 1}}{{x + 1}}} \right) = 1\)

\( \Leftrightarrow \mathop {\lim }\limits_{x \to  + \infty } \left( {\frac{{\left( {a + 1} \right)x + \left( {a + b + 3} \right) + \frac{{b + 1}}{x}}}{{1 + \frac{1}{x}}}} \right) = 1\) \( \Leftrightarrow \left\{ \begin{array}{l}a + 1 = 0\\a + b + 3 = 1\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}a =  - 1\\b =  - 1\end{array} \right.\).

Do đó T = 2024a – 4049b = 2024.(−1) – 4049.(−1) = 2025.

Trả lời: 2025.