Giải thích các bước giải:
\(\begin{array}{l}
d,\\
\lim \left( {\sqrt {3{n^2} + 2n - 1} - \sqrt {3{n^2} - 4n + 8} } \right)\\
= \lim \frac{{\left( {\sqrt {3{n^2} + 2n - 1} - \sqrt {3{n^2} - 4n + 8} } \right)\left( {\sqrt {3{n^2} + 2n - 1} + \sqrt {3{n^2} - 4n + 8} } \right)}}{{\sqrt {3{n^2} + 2n - 1} + \sqrt {3{n^2} - 4n + 8} }}\\
= \lim \frac{{\left( {3{n^2} + 2n - 1} \right) - \left( {3{n^2} - 4n + 8} \right)}}{{\sqrt {3{n^2} + 2n - 1} + \sqrt {3{n^2} - 4n + 8} }}\\
= \lim \frac{{6n - 9}}{{\sqrt {3{n^2} + 2n - 1} + \sqrt {3{n^2} - 4n + 8} }}\\
= \lim \frac{{6 - \frac{9}{n}}}{{\sqrt {3 + \frac{2}{n} - \frac{1}{{{n^2}}}} + \sqrt {3 - \frac{4}{n} + \frac{8}{{{n^2}}}} }}\\
= \frac{6}{{\sqrt 3 + \sqrt 3 }} = \sqrt 3 \\
e,\\
\lim \frac{{\sqrt {{n^2} + 1} - \sqrt {n + 1} }}{{3n + 2}}\\
= \lim \frac{{\sqrt {1 + \frac{1}{{{n^2}}}} - \sqrt {\frac{1}{n} + \frac{1}{{{n^2}}}} }}{{3 + \frac{2}{n}}}\\
= \frac{{\sqrt 1 - \sqrt 0 }}{3}\\
= \frac{1}{3}\\
f,\\
\lim \frac{{\sqrt {{n^2} + n - 1} - \sqrt {4{n^2} - 2} }}{{n + 3}}\\
= \lim \frac{{\sqrt {1 + \frac{1}{n} - \frac{1}{{{n^2}}}} - \sqrt {4 - \frac{2}{{{n^2}}}} }}{{1 + \frac{3}{n}}}\\
= \frac{{\sqrt 1 - \sqrt 4 }}{1}\\
= - 1
\end{array}\)
