Đáp án:
18C
Giải thích các bước giải:
\(\begin{array}{l}
C18:\\
\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^2} + 3} - x - 4}}{{\sqrt {{x^2} + 2} - x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - \sqrt {1 + \dfrac{3}{{{x^2}}}} - 1 - \dfrac{4}{x}}}{{ - \sqrt {1 + \dfrac{2}{{{x^2}}}} - 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - 1 - 1}}{{ - 1 - 1}} = 1\\
\to C\\
C19:\\
\lim \dfrac{{\sqrt {{n^2} + 1} - \sqrt[3]{{4{n^3} + 2}}}}{{\sqrt[4]{{2{n^4} + n + 2}} - n}}\\
= \lim \dfrac{{\sqrt {1 + \dfrac{1}{{{n^2}}}} - \sqrt[3]{{4 + \dfrac{2}{{{n^3}}}}}}}{{\sqrt[4]{{2 + \dfrac{1}{{{n^3}}} + \dfrac{2}{{{n^4}}}}} - 1}} = \dfrac{{1 - \sqrt[3]{4}}}{{\sqrt[4]{2} - 1}}\\
\to a = 4;b = 2\\
\to a.b = 4.2 = 8\\
\to B
\end{array}\)
