Đáp án:
$\begin{array}{l}
C1)\\
L = \lim \dfrac{{n\sqrt n + 1}}{{3{n^2} + 2}}\\
= \lim \dfrac{{\dfrac{{n\sqrt n }}{{{n^2}}} + \dfrac{1}{{{n^2}}}}}{{\dfrac{{3{n^2}}}{3} + \dfrac{2}{{{n^2}}}}}\\
= \lim \dfrac{{\dfrac{1}{{\sqrt n }} + \dfrac{1}{{{n^2}}}}}{{3 + \dfrac{2}{{{n^2}}}}}\\
Do:\lim \left( {\dfrac{1}{{\sqrt n }} + \dfrac{1}{{{n^2}}}} \right) = 0\\
\lim \left( {3 + \dfrac{2}{{{n^2}}}} \right) = 3\\
\Rightarrow \lim \dfrac{{\dfrac{1}{{\sqrt n }} + \dfrac{1}{{{n^2}}}}}{{3 + \dfrac{2}{{{n^2}}}}} = \dfrac{0}{3} = 0\\
C2)\\
L = \lim \dfrac{{\dfrac{{n\sqrt n }}{{n\sqrt n }} + \dfrac{1}{{n\sqrt n }}}}{{\dfrac{{3{n^2}}}{{n\sqrt n }} + \dfrac{2}{{n\sqrt n }}}}\\
= \lim \dfrac{{1 + \dfrac{1}{{n\sqrt n }}}}{{3.\sqrt n + \dfrac{2}{{n\sqrt n }}}}\\
Do:\left\{ \begin{array}{l}
\lim \left( {1 + \dfrac{1}{{n\sqrt n }}} \right) = 1\\
\lim \left( {3\sqrt n + \dfrac{2}{{n\sqrt n }}} \right) = \infty
\end{array} \right.\\
\Rightarrow L = \dfrac{1}{\infty } = 0
\end{array}$
