Đáp án:
$\begin{array}{l}
b)\lim \dfrac{{n + 3}}{{n{{.4}^n}}}\\
= \lim \dfrac{{\dfrac{n}{{n{{.4}^n}}} + \dfrac{3}{{n{{.4}^n}}}}}{1}\\
= \lim \left( {\dfrac{1}{{{4^n}}} + \dfrac{3}{{n{{.4}^n}}}} \right)\\
= 0\\
c)lim\dfrac{{{1^2} + {2^2} + {3^2} + ... + {n^2}}}{{2n\sqrt {1 + 3 + 5 + ... + \left( {2n - 1} \right)} }}\\
= \lim \dfrac{{{1^2} + {2^2} + {3^2} + ... + {n^2}}}{{2n.\sqrt {\dfrac{{\left( {2n - 1 + 1} \right).n}}{2}} }}\\
= \lim \dfrac{{{1^2} + {2^2} + {3^2} + ... + {n^2}}}{{2n.\sqrt {{n^2}} }}\\
= \lim \dfrac{{{1^2} + {2^2} + {3^2} + ... + {n^2}}}{{2n.n}}\\
= \lim \dfrac{{\dfrac{{{1^2} + {2^2} + {3^2} + ... + {n^2}}}{{{n^2}}}}}{2}\\
= \lim \dfrac{{\dfrac{{{1^2} + {2^2} + {3^2} + ... + {{\left( {n - 1} \right)}^2}}}{{{n^2}}} + 1}}{2}\\
= \dfrac{1}{2}
\end{array}$
