Giải thích các bước giải:
\(\begin{array}{l}
a,\\
\lim \left( {\sqrt {4{n^2} - 3n + 5} - 4n + 1} \right)\\
= \lim \frac{{\left( {4{n^2} - 3n + 5} \right) - {{\left( {4n - 1} \right)}^2}}}{{\sqrt {4{n^2} - 3n + 5} + 4n - 1}}\\
= \lim \frac{{4{n^2} - 3n + 5 - 16{n^2} + 8n - 1}}{{\sqrt {4{n^2} - 3n + 5} + 4n - 1}}\\
= \lim \frac{{ - 12{n^2} + 5n + 4}}{{\sqrt {4{n^2} - 3n + 5} + 4n - 1}}\\
= \lim \frac{{ - 12n + 5 - \frac{4}{n}}}{{\sqrt {4 - \frac{3}{n} + \frac{5}{{{n^2}}}} + 4 - \frac{1}{n}}}\\
= \frac{{ - \infty }}{{\sqrt 4 + 4}} = - \infty \\
b,\\
\lim \left( {2n + 1 - \sqrt {2{n^2} - n + 1} } \right)\\
= \lim \frac{{{{\left( {2n + 1} \right)}^2} - \left( {2{n^2} - n + 1} \right)}}{{2n + 1 + \sqrt {2{n^2} - n + 1} }}\\
= \lim \frac{{4{n^2} + 4n + 1 - 2{n^2} + n - 1}}{{2n + 1 + \sqrt {2{n^2} - n + 1} }}\\
= \lim \frac{{2{n^2} + 5n}}{{2n + 1 + \sqrt {2{n^2} - n + 1} }}\\
= \lim \frac{{2n + 5}}{{2 + \frac{1}{n} + \sqrt {2 - \frac{1}{n} + \frac{1}{{{n^2}}}} }}\\
= \frac{{ + \infty }}{{2 + \sqrt 2 }} = + \infty \\
c,\\
\lim \frac{{ - {n^2} - 4n + 1}}{{3 + 2n}}\\
= \lim \frac{{ - n - 4 + \frac{1}{n}}}{{\frac{3}{n} + 2}}\\
= \frac{{ - \infty }}{2} = - \infty \\
d,\\
\lim \frac{{3{n^3} - 4n + 2}}{{2{n^2} + n + 1}}\\
= \lim \frac{{3n - \frac{4}{n} + \frac{2}{{{n^2}}}}}{{2 + \frac{1}{n} + \frac{1}{{{n^2}}}}}\\
= \frac{{ + \infty }}{2} = + \infty \\
e,\\
\lim \frac{{1 - 4n - {n^2}}}{{2n + 1}}\\
= \lim \frac{{\frac{1}{n} - 4 - n}}{{2 + \frac{1}{n}}}\\
= \frac{{ - \infty }}{2}\\
= - \infty
\end{array}\)
