Giải thích các bước giải:
$\begin{array}{l}
f)\lim \left( {\sqrt {{n^2} + 3n} - n + 2} \right)\\
= \lim \dfrac{{{n^2} + 3n - {{\left( {n - 2} \right)}^2}}}{{\sqrt {{n^2} + 3n} + n - 2}}\\
= \lim \dfrac{{{n^2} + 3n - {n^2} + 4n - 4}}{{\sqrt {{n^2} + 3n} + n - 2}}\\
= \lim \dfrac{{7n - 4}}{{\sqrt {{n^2} + 3n} + n - 2}}\\
= \lim \dfrac{{7 - \dfrac{4}{n}}}{{\sqrt {1 + \dfrac{3}{n}} + 1 - \dfrac{2}{n}}}\\
= \dfrac{{7 - 0}}{{\sqrt {1 + 0} + 1 - 0}}\\
= \dfrac{7}{2}\\
e)\lim \left( {\sqrt {n + 1} - \sqrt n } \right)\\
= \lim \dfrac{{n + 1 - n}}{{\sqrt {n + 1} + \sqrt n }}\\
= \lim \dfrac{1}{{\sqrt {n + 1} + \sqrt n }}\\
= 0\\
h)\lim \dfrac{{\sqrt {4{n^2} + 1} - 2n + 1}}{{\sqrt {{n^2} + 2n} - n}}\\
= \lim \dfrac{{4{n^2} + 1 - {{\left( {2n - 1} \right)}^2}}}{{\sqrt {4{n^2} + 1} + 2n - 1}}.\dfrac{{\sqrt {{n^2} + 2n} + n}}{{{n^2} + 2n - {n^2}}}\\
= \lim \dfrac{{4{n^2} + 1 - {{\left( {2n - 1} \right)}^2}}}{{\sqrt {4{n^2} + 1} + 2n - 1}}.\lim \dfrac{{\sqrt {{n^2} + 2n} + n}}{{{n^2} + 2n - {n^2}}}\\
= \lim \dfrac{{4n}}{{\sqrt {4{n^2} + 1} + 2n - 1}}.\lim \dfrac{{\sqrt {{n^2} + 2n} + n}}{{2n}}\\
= \lim \dfrac{4}{{\sqrt {4 + \dfrac{1}{{{n^2}}}} + 2}}.\lim \dfrac{{\sqrt {1 + \dfrac{2}{n}} + 1}}{2}\\
= \dfrac{4}{{\sqrt {4 + 0} + 2}}.\dfrac{{\sqrt {1 + 0} + 1}}{2}\\
= 1
\end{array}$
