Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
E = \lim \frac{{\sqrt {{n^3} + 2n} + 1}}{{n + 2}}\\
= \lim \frac{{\frac{{\sqrt {{n^3} + 2n} + 1}}{n}}}{{\frac{{n + 2}}{n}}}\\
= \lim \frac{{\sqrt {n + \frac{2}{n}} + \frac{1}{n}}}{{1 + \frac{2}{n}}}\\
\lim \left( {\sqrt {n + \frac{2}{n}} + \frac{1}{n}} \right) = + \infty \\
\lim \left( {1 + \frac{2}{n}} \right) = 1\\
\Rightarrow E = \lim \frac{{\sqrt {n + \frac{2}{n}} + \frac{1}{n}}}{{1 + \frac{2}{n}}} = + \infty \\
2,\\
\lim \sqrt {3 + \frac{{{n^2} - 1}}{{3 + {n^2}}} - \frac{1}{{{2^n}}}} = \lim \sqrt {3 + \frac{{1 - \frac{1}{{{n^2}}}}}{{\frac{3}{{{n^2}}} + 1}} - {{\left( {\frac{1}{2}} \right)}^n}} \\
\lim \frac{{1 - \frac{1}{{{n^2}}}}}{{\frac{3}{{{n^2}}} + 1}} = \frac{1}{1} = 1\\
\lim {\left( {\frac{1}{2}} \right)^n} = 0\\
\Rightarrow \lim \sqrt {3 + \frac{{{n^2} - 1}}{{3 + {n^2}}} - \frac{1}{{{2^n}}}} = \sqrt {3 + 1 - 0} = 2
\end{array}\)
