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. $
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. $