Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \dfrac{{\left( {1 - 2n} \right)\sqrt {3{n^2} + 2} }}{{4n + 5}}\\
= \lim \dfrac{{\left( {\frac{1}{n} - 2} \right)\sqrt {3{n^2} + 2} }}{{4 + \frac{5}{n}}}\\
= \lim \left[ {\sqrt {3{n^2} + 2} .\dfrac{{\frac{1}{n} - 2}}{{4 + \frac{5}{n}}}} \right]\\
\lim \left( {3{n^2} + 2} \right) = + \infty \Rightarrow \lim \sqrt {3{n^2} + 2} = + \infty \\
\lim \frac{{\frac{1}{n} - 2}}{{4 + \frac{5}{n}}} = \frac{{0 - 2}}{{4 + 0}} = - \frac{1}{2}\\
\Rightarrow \lim \left[ {\sqrt {3{n^2} + 2} .\dfrac{{\frac{1}{n} - 2}}{{4 + \frac{5}{n}}}} \right] = - \infty \\
\Rightarrow \lim \frac{{\left( {1 - 2n} \right)\sqrt {3{n^2} + 2} }}{{4n + 5}} = - \infty \\
\end{array}\)
