Đáp án:
\(\begin{array}{l}
B2:\\
a)0\\
b)0\\
B3:\\
- \dfrac{1}{4}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
a)\lim \left( {n + 1} \right).\sqrt {\dfrac{{n - 2}}{{{n^4} + 2{n^2} + 1}}} \\
= \lim \sqrt {\dfrac{{{{\left( {n + 1} \right)}^2}\left( {n - 2} \right)}}{{{n^4} + 2{n^2} + 1}}} \\
= \lim \sqrt {\dfrac{{{{\left( {1 + \dfrac{1}{n}} \right)}^2}.\left( {\dfrac{1}{n} - \dfrac{2}{{{n^2}}}} \right)}}{{1 + \dfrac{2}{{{n^2}}} + \dfrac{1}{{{n^4}}}}}} \\
= \lim \dfrac{0}{1} = 0\\
b)\lim \left( {\sqrt {{n^2} + 3} - n} \right)\\
= \lim \dfrac{{{n^2} + 3 - {n^2}}}{{\sqrt {{n^2} + 3} + n}}\\
= \lim \dfrac{3}{{\sqrt {{n^2} + 3} + n}}\\
= \lim \dfrac{{\dfrac{3}{n}}}{{\sqrt {1 + \dfrac{3}{{{n^2}}}} + 1}} = \dfrac{0}{2} = 0
\end{array}\)
