Mn giúp em với ạ . Em cảm ơn

Đáp án:

 f) 2

Giải thích các bước giải:

\(\begin{array}{l}
C1:\\
a)\lim \dfrac{{\dfrac{1}{n}}}{{1 + \dfrac{2}{n}}} = \dfrac{0}{1} = 0\\
b)\lim \dfrac{{\dfrac{1}{{{n^k}}}}}{1} = \dfrac{0}{1} = 0\\
g)\lim \dfrac{{5 + \dfrac{1}{{{n^3}}}}}{{1.{{\left( {1 + \dfrac{1}{n}} \right)}^2}}} = \dfrac{5}{1} = 5\\
c)\lim \dfrac{{6 - \dfrac{1}{n}}}{{\sqrt {1 + \dfrac{4}{n} + \dfrac{2}{{{n^2}}}} }} = \dfrac{6}{1} = 6\\
d)\lim \dfrac{{5 + \dfrac{1}{n}}}{{\dfrac{1}{n} + 3}} = \dfrac{5}{3}\\
h)\lim \dfrac{{\dfrac{1}{n} - \dfrac{3}{{{n^2}}} + \dfrac{2}{{{n^4}}}}}{{3 - \dfrac{1}{n} + \dfrac{1}{{{n^4}}}}} = \dfrac{0}{3} = 0\\
e)\lim \dfrac{{4 + \dfrac{3}{n} + \dfrac{1}{{{n^2}}}}}{{{{\left( {1 - \dfrac{1}{n}} \right)}^2}}} = \dfrac{4}{1} = 4\\
f)\lim \dfrac{{{{\left( {2 + \dfrac{1}{{{n^2}}}} \right)}^3}.{{\left( {1 + \dfrac{2}{n}} \right)}^5}}}{{4 - \dfrac{1}{{{n^{11}}}}}} = \dfrac{{{2^3}{{.1}^5}}}{4} = 2\\
i)\lim \dfrac{{\dfrac{n}{{{4^n}}} + \dfrac{5}{{{4^n}}}}}{n} = \dfrac{0}{n} = 0
\end{array}\)