Đáp án:
4B
Giải thích các bước giải:
\(\begin{array}{l}
a)\lim \dfrac{{\dfrac{{2{n^2}}}{{{n^2}{{.5}^n}}} + \dfrac{3}{{{n^2}{{.5}^n}}}}}{1} = \dfrac{0}{1} = 0\\
l)\lim \dfrac{{{n^3} + {n^2} - {n^3}}}{{\sqrt[3]{{{{\left( {{n^3} + {n^2}} \right)}^2}}} + n\sqrt[3]{{{n^3} + {n^2}}} + {n^2}}}\\
= \lim \dfrac{{{n^2}}}{{\sqrt[3]{{{{\left( {{n^3} + {n^2}} \right)}^2}}} + n\sqrt[3]{{{n^3} + {n^2}}} + {n^2}}}\\
= \lim \dfrac{1}{{\sqrt[3]{{{{\left( {1 + \dfrac{1}{n}} \right)}^2}}} + 1.\sqrt[3]{{1 + \dfrac{1}{n}}} + 1}}\\
= \dfrac{1}{{1 + 1 + 1}} = \dfrac{1}{3}\\
C4:\\
\lim \dfrac{{\dfrac{1}{{ - {n^4}{{.8}^n}}} + \dfrac{3}{{{8^n}}}}}{1} = \dfrac{0}{1} = 0\\
\to B\\
C5:\\
\lim n\left( {2 - \dfrac{3}{n}} \right) = + \infty \\
Do:\mathop {\lim }\limits_{x \to + \infty } n = + \infty \\
\mathop {\lim }\limits_{x \to + \infty } \left( {2 - \dfrac{3}{n}} \right) = 2
\end{array}\)
