Đáp án:
$D$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\lim \dfrac{{a{n^2} + \sqrt {{n^2} + 3n - 7} }}{{bn - 3}} = 2\\
\Leftrightarrow \left\{ \begin{array}{l}
a = 0\\
\lim \dfrac{{\sqrt {{n^2} + 3n - 7} }}{{bn - 3}} = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = 0\\
\lim \dfrac{{\sqrt {1 + \dfrac{3}{{{n^2}}} - \dfrac{7}{{{n^2}}}} }}{{b - \dfrac{3}{n}}} = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = 0\\
\dfrac{1}{b} = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = 0\\
b = \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow a + 4b = 2
\end{array}$
