Đáp án:
1) A
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
\lim \dfrac{{1 + \dfrac{4}{{{n^2}}} - \dfrac{5}{{{n^3}}}}}{{3 + \dfrac{1}{n} + \dfrac{7}{{{n^3}}}}} = \dfrac{1}{3}\\
\to A\\
C2:\\
\lim \dfrac{{\dfrac{2}{{{5^n}}} - {5^2}}}{{{{\left( {\dfrac{3}{5}} \right)}^n} + 2}} = - \dfrac{{25}}{2}\\
\to A\\
C3:\\
\lim \dfrac{{n + 1 - n}}{{\sqrt {n + 1} + \sqrt n }} = \lim \dfrac{1}{{\sqrt {n + 1} + \sqrt n }}\\
= \lim \dfrac{{\dfrac{1}{{\sqrt n }}}}{{\sqrt {1 + \dfrac{1}{n}} + \sqrt 1 }} = 0\\
\to D\\
C4:\\
\lim {n^3}\left( {\dfrac{5}{{{n^2}}} - 3} \right) = - \infty \\
Do:\mathop {\lim }\limits_{x \to + \infty } {n^3} = + \infty \\
\mathop {\lim }\limits_{x \to + \infty } \left( {\dfrac{5}{{{n^2}}} - 3} \right) = - 3\\
\to B\\
C5:\\
\lim \dfrac{{{n^2} - 2n}}{{5n + 5{n^2}}} = \lim \dfrac{{1 - \dfrac{2}{n}}}{{\dfrac{5}{n} + 5}} = \dfrac{1}{5}\\
\to C\\
C6:\\
\lim \dfrac{{{{\left( {\dfrac{2}{5}} \right)}^n} + 1}}{1} = 1\\
\to A\\
C7:\\
\lim \dfrac{{\dfrac{1}{n} - 4}}{5} = - \dfrac{4}{5}\\
\to D\\
C8:\\
\lim \dfrac{{\sqrt {9 - \dfrac{1}{n} + \dfrac{1}{{{n^2}}}} }}{{4 - \dfrac{2}{n}}} = \dfrac{3}{4}\\
\to D\\
C9:D\\
C10:D\\
C11:\\
\lim {n^2}\left( {3 + \dfrac{5}{n} - \dfrac{3}{{{n^2}}}} \right) = + \infty \\
Do:\mathop {\lim }\limits_{x \to + \infty } {n^2} = + \infty \\
\lim \left( {3 + \dfrac{5}{n} - \dfrac{3}{{{n^2}}}} \right) = 3\\
\to C
\end{array}\)
