a. Áp dụng BĐT |a+b|≤|a|+|b|. Ta có$|(-1)^n\sin n^2+\cos n| \le |(-1)^n\sin n^2| +|\cos n| \le 1+1=2.Suyra-\frac{2}{2\sqrt[3]{n}+1} \le \frac{(-1)^n\sin n^2+\cos n}{2\sqrt[3]{n}+1} \le \frac{2}{2\sqrt[3]{n}+1}.Mặt khác\lim \frac{2}{2\sqrt[3]{n}+1} = \lim \left (- \frac{2}{2\sqrt[3]{n}+1} \right )=0.Vậy \lim\frac{(-1)^n\sin n^2+\cos n}{2\sqrt[3]{n}+1}=0. $
a. Ta có$|(-1)^n\sin n^2+\cos n| \le |(-1)^n\sin n^2| +|\cos n| =1+1=2. Suy ra -\frac{2}{2\sqrt[3]{n}+1} \le \frac{(-1)^n\sin n^2+\cos n}{2\sqrt[3]{n}+1} \le \frac{2}{2\sqrt[3]{n}+1}.Mặt khác\lim \frac{2}{2\sqrt[3]{n}+1} = \lim \left (- \frac{2}{2\sqrt[3]{n}+1} \right )=0.Vậy \lim\frac{(-1)^n\sin n^2+\cos n}{2\sqrt[3]{n}+1}=0. $
a.
Áp dụng BĐT |a+b| \le |a|+|b|. Ta có$|(-1)^n\sin n^2+\cos n| \le |(-1)^n\sin n^2| +|\cos n|
\le 1+1=2
. Suy ra -\frac{2}{2\sqrt[3]{n}+1} \le \frac{(-1)^n\sin n^2+\cos n}{2\sqrt[3]{n}+1} \le \frac{2}{2\sqrt[3]{n}+1}
.Mặt khác\lim \frac{2}{2\sqrt[3]{n}+1} = \lim \left (- \frac{2}{2\sqrt[3]{n}+1} \right )=0
.Vậy \lim\frac{(-1)^n\sin n^2+\cos n}{2\sqrt[3]{n}+1}=0. $