Đáp án:
$\begin{array}{l}
a)\lim \dfrac{{\left( {n - 1} \right)\left( {n - 2{n^2}} \right)}}{{3{n^3} - 4n + 1}}\\
= \lim \dfrac{{{n^2} - 2{n^3} - n + 2{n^2}}}{{3{n^3} - 4n + 1}}\\
= \lim \dfrac{{ - 2{n^3} + 3{n^2} - n}}{{3{n^3} - 4n + 1}}\\
= \lim \dfrac{{ - 2 + \dfrac{3}{n} - \dfrac{1}{{{n^2}}}}}{{3 - \dfrac{4}{{{n^2}}} + \dfrac{1}{{{n^3}}}}}\\
= \dfrac{{ - 2}}{3}\\
b)\lim \dfrac{{{3^n} - {2^n}}}{{{{4.3}^{n + 1}} - 3}}\\
= \lim \dfrac{{{3^n} - {2^n}}}{{{{12.3}^n} - 3}}\\
= \lim \dfrac{{1 - {{\left( {\dfrac{2}{3}} \right)}^n}}}{{12 - \dfrac{3}{{{3^n}}}}}\\
= \dfrac{1}{{12}}\\
c)\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 4}}{{{x^2} - 3x + 2}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{x + 2}}{{x - 1}}\\
= \dfrac{{2 + 2}}{{2 - 1}}\\
= 4
\end{array}$
