Giải thích các bước giải:
\(\begin{array}{l}
a,\\
\lim \frac{{{n^2} + 2n - 4}}{{7{n^3} - 2n + 9}}\\
= \lim \dfrac{{\frac{{{n^2} + 2n - 4}}{{{n^3}}}}}{{\frac{{7{n^3} - 2n + 9}}{{{n^3}}}}}\\
= \lim \dfrac{{\frac{1}{n} + \frac{2}{{{n^2}}} - \frac{4}{{{n^3}}}}}{{7 - \frac{2}{{{n^2}}} + \frac{9}{{{n^3}}}}}\\
= \frac{0}{7} = 0\\
b,\\
\lim \frac{{\sqrt {{n^2} + 2} }}{{\sqrt {4{n^2} - 2} }}\\
= \lim \dfrac{{\frac{{\sqrt {{n^2} + 2} }}{n}}}{{\frac{{\sqrt {4{n^2} - 2} }}{n}}}\\
= \lim \frac{{\sqrt {1 + \frac{2}{{{n^2}}}} }}{{\sqrt {4 - \frac{2}{{{n^2}}}} }}\\
= \frac{{\sqrt 1 }}{{\sqrt 4 }} = \frac{1}{2}\\
c,\\
\lim \left( {\sqrt {{n^2} + 2n - 3} - n} \right)\\
= \lim \frac{{\left( {{n^2} + 2n - 3} \right) - {n^2}}}{{\sqrt {{n^2} + 2n - 3} + n}}\\
= \lim \frac{{2n - 3}}{{\sqrt {{n^2} + 2n - 3} + n}}\\
= \lim \frac{{2 - \frac{3}{n}}}{{\sqrt {1 + \frac{2}{n} - \frac{3}{{{n^2}}}} + 1}}\\
= \frac{2}{{\sqrt 1 + 1}}\\
= 1\\
d,\\
\lim \left( {\sqrt {n - 1} - \sqrt n } \right)\\
= \lim \frac{{\left( {n - 1} \right) - n}}{{\sqrt {n - 1} + \sqrt n }}\\
= \lim \frac{{ - 1}}{{\sqrt {n - 1} + \sqrt n }}\\
= 0\\
\left( {{\rm{do}}\,\,\,{\rm{lim}}\left( {\sqrt {n - 1} + \sqrt n } \right) = + \infty } \right)\\
e,\\
\lim \frac{{{5^n} - {4^n}}}{{{{3.5}^n} + 1}}\\
= \lim \frac{{1 - {{\left( {\frac{4}{5}} \right)}^n}}}{{3 + {{\left( {\frac{1}{5}} \right)}^n}}}\\
= \frac{1}{3}
\end{array}\)
