Đáp án:
\(\lim \frac{{\left( {1 - 2n} \right)\left( {3 + 2n} \right)}}{{{n^2} - n + 1}} - 4\)
\(\lim \frac{{\sqrt {{n^2} + 2} }}{{2 - n}} = - 1\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\lim \frac{{\left( {1 - 2n} \right)\left( {3 + 2n} \right)}}{{{n^2} - n + 1}}\\
= \lim \frac{{\left( {\frac{1}{n} - 2} \right)\left( {\frac{3}{n} + 2} \right)}}{{1 - \frac{1}{n} + \frac{1}{{{n^2}}}}}\\
= \frac{{ - 2.2}}{1} = - 4\\
b.\lim \frac{{\sqrt {{n^2} + 2} }}{{2 - n}}\\
= \lim \frac{{\sqrt {1 + \frac{2}{{{n^2}}}} }}{{\frac{2}{n} - 1}} = - 1
\end{array}\)
