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

a) \(\mathop {\lim }\limits_{x \to - 2} f\left( x \right) = \sqrt 5 + m\).

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

c) \(\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = \sqrt 2 + m\).

d) Khi \(m = 3 + \sqrt 2 \) thì hàm số đã cho có giới hạn tại x₀ = −1.

\(\mathop {\lim }\limits_{x \to 3} \left[ {4x - 3f\left( x \right)} \right]\)\( = \mathop {\lim }\limits_{x \to 3} \left( {4x} \right) - 3\mathop {\lim }\limits_{x \to 3} f\left( x \right) = 12 - 3.2 = 6\).

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

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

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

Trả lời: 2025.