Kết quả giới hạn \(\mathop {\lim }\limits_{x \to - \infty } \frac{{1 + 3x}}{{\sqrt {2{x^2} + 3} }}\) là

a) \(\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {{n^2} + 3n + 1} - n} \right) = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2} + 3n + 1 - {n^2}}}{{\sqrt {{n^2} + 3n + 1} + n}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{3n + 1}}{{\sqrt {{n^2} + 3n + 1} + n}}\)

\( = \mathop {\lim }\limits_{n \to + \infty } \frac{{3 + \frac{1}{n}}}{{\sqrt {1 + \frac{3}{n} + \frac{1}{{{n^2}}}} + 1}} = \frac{3}{2}\)

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

\( = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 1} \right)}}{{ - \left( {x + 2} \right)}} = - \frac{1}{4}\)

Chọn C

Nếu \(f\left( 2 \right) \ne - 1 \Rightarrow \mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) + 1}}{{x - 2}} = \infty \) ( mâu thuẫn giả thiết )

Do đó \(f\left( 2 \right) = - 1\)

Ta có \(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {f(x) + 2x + 1} - x}}{{{x^2} - 4}} = T\)và ta có

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {f(x) + 2x + 1} - x}}{{{x^2} - 4}}\\ = \mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) + 1 + 2x - {x^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {\left[ {\sqrt {f(x) + 2x + 1} + x} \right]} \right)}}\\ = \mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) + 1}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {\left[ {\sqrt {f(x) + 2x + 1} + x} \right]} \right)}} + \mathop {\lim }\limits_{x \to 2} \frac{{ - x\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {\left[ {\sqrt {f(x) + 2x + 1} + x} \right]} \right)}}\\ = \frac{a}{{4.\left( {2 + 2} \right)}} + \mathop {\lim }\limits_{x \to 2} \frac{{ - x}}{{\left( {x + 2} \right)\left( {\left[ {\sqrt {f(x) + 2x + 1} + x} \right]} \right)}} = \frac{a}{{16}} - \frac{2}{{4\left( {2 + 2} \right)}} = \frac{a}{{16}} - \frac{1}{8} = \frac{{a - 2}}{{16}}\end{array}\)

Hay là \(T = \frac{{a - 2}}{{16}}\).