\(\lim\dfrac{6^n.cos3n+5^n}{2^n+2.7^n}=\lim\dfrac{7^n.\left[\left(\dfrac{6}{7}\right)^n.cos3x+\left(\dfrac{5}{7}\right)^n\right]}{7^n.\left[\left(\dfrac{2}{7}\right)^n+1\right]}=\lim\dfrac{\left(\dfrac{6}{7}\right)^n.cos3n+\left(\dfrac{5}{7}\right)^n}{\left(\dfrac{2}{7}\right)^n+1}\)
Do \(\left(\dfrac{6}{7}\right)^n\rightarrow0\) và \(-1\le cos3n\le1\Rightarrow\left(\dfrac{6}{7}\right)^n.cos3n\rightarrow0\)
\(\Rightarrow\lim\dfrac{6^n.cos3n+5^n}{2^n+2.7^n}=\dfrac{0+0}{0+1}=0\)
\(\lim\dfrac{4n.sin^n2n+cos^n2n}{4n^2+8n}=\lim\dfrac{n^2\left(\dfrac{4.sin^n2n}{n}+\dfrac{cos^n2n}{n}\right)}{n^2\left(4+\dfrac{8}{n}\right)}=\lim\dfrac{\dfrac{4.sin^n2n}{n}+\dfrac{cos^n2n}{n}}{4+\dfrac{8}{n}}\)
Do \(\dfrac{1}{n}\rightarrow0\) và \(\left\{{}\begin{matrix}-1\le sin^n2n\le1\\-1\le cos^n2n\le1\end{matrix}\right.\) \(\Rightarrow\dfrac{4sin^n2n}{n}+\dfrac{cos^n2n}{n}\rightarrow0\)
\(\Rightarrow\lim...=\dfrac{0+0}{4+0}=0\)
