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