Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \lim \frac{{3n - 1}}{{2n + 2}} = \lim \frac{{3 - \frac{1}{n}}}{{2 + \frac{2}{n}}} = \frac{3}{2}\\
H = \lim \left( {\sqrt {{n^2} - 2n} - n} \right) = \lim \frac{{{n^2} - 2n - {n^2}}}{{\sqrt {{n^2} - 2n} + n}}\\
= \lim \frac{{ - 2n}}{{\sqrt {{n^2} - 2n} + n}}\\
= \lim \frac{{ - 2}}{{\sqrt {1 - \frac{2}{n}} + 1}} = - \frac{2}{{1 + 1}} = - 1\\
N = \lim \frac{{\sqrt n - 2}}{{3n + 7}} = \lim \frac{{\frac{1}{{\sqrt n }} - \frac{2}{n}}}{{3 + \frac{7}{n}}} = 0\\
B = \lim \frac{{{3^n} - 5 + {4^n}}}{{1 - {4^n}}} = \lim \frac{{{{\left( {\frac{3}{4}} \right)}^n} - \frac{5}{{{4^n}}} + 1}}{{\frac{1}{{{4^n}}} - 1}} = \frac{1}{{ - 1}} = - 1
\end{array}\)
