Tìm: a) L = lim $\frac{(2n^{2}-n)^5.(4n-1)^4}{20n^6.(2n^2-n+1)^2}$ b) L = lim $\frac{2n^2-11n+1}{n^2-1}$ Mn giúp mik vs 🙏

Đáp án:

$\begin{array}{l}
a)L = \lim \dfrac{{{{\left( {2{n^2} - n} \right)}^5}.{{\left( {4n - 1} \right)}^4}}}{{20{n^6}.{{\left( {2{n^2} - n + 1} \right)}^2}}}\\
 = \lim \dfrac{{\dfrac{1}{n}.\dfrac{{{{\left( {2{n^2} - n} \right)}^5}}}{{{n^5}}}.\dfrac{{{{\left( {4n - 1} \right)}^4}}}{{{n^4}}}}}{{\dfrac{{20{n^6}}}{{{n^6}}}.\dfrac{{{{\left( {2{n^2} - n + 1} \right)}^2}}}{{{n^4}}}}}\\
 = \lim \dfrac{{\dfrac{1}{n}.{{\left( {2n - 1} \right)}^5}.{{\left( {4 - \dfrac{1}{n}} \right)}^4}}}{{20.{{\left( {2 - \dfrac{1}{n} + \dfrac{1}{{{n^2}}}} \right)}^2}}}\\
 = 0\\
b)L = \lim \dfrac{{2{n^2} - 11n + 1}}{{{n^2} - 1}}\\
 = \lim \dfrac{{2 - \dfrac{{11}}{n} + \dfrac{1}{{{n^2}}}}}{{1 - \dfrac{1}{{{n^2}}}}}\\
 = \dfrac{2}{1}\\
 = 2
\end{array}$