Tính giới hạn

A

- Hướng dẫn giải

Đặt $u_n=\sqrt[3]{n^3+3n^2+n+1}$

Ta có 

$$\begin{gathered} \cos \left({\mathrm{\pi nu}}_{\mathrm{n}}\right) \\ =\cos \left[-{\mathrm{\pi nu}}_{\mathrm{n}}+\left(\mathrm{n}+1\right)\mathrm{n\pi }\right] \\ =\mathrm{cos\pi }n\left(n+1-u_n\right) \\ =\cos \left[\mathrm{\pi n}\frac{{\left(\mathrm{n}+1\right)}^3-{{\mathrm{u}}^3}_{\mathrm{n}}}{{\left(\mathrm{n}+1\right)}^2+\left(\mathrm{n}+1\right){\mathrm{u}}_{\mathrm{n}}}\right] \\ =\cos \left[\frac{2{\mathrm{\pi u}}^2}{{\left(\mathrm{n}+1\right)}^2+\left(\mathrm{n}+1\right){\mathrm{u}}_{\mathrm{n}}+{{\mathrm{u}}^2}_{\mathrm{n}}}\right] \\ =\cos \left[\frac{2\mathrm{\pi }}{{\left(1+{\displaystyle \frac{1}{n}}\right)}^2+{\displaystyle \frac{u_n}{n}}\left(1+\frac{1}{n}\right)}\right] \end{gathered}$$

Suy ra 

$$\begin{gathered} \underset{n\to \infty }{\lim }\cos \left({\mathrm{\pi nu}}_{\mathrm{n}}\right) \\ =\cos \left(\frac{2\mathrm{\pi }}{3}\right)=-\frac{1}{2} \end{gathered}$$

Biến đổi tương tự, ta cũng tìm được 

$$\begin{gathered} \underset{n\to \infty }{\lim }\sin \left({\mathrm{\pi nu}}_{\mathrm{n}}\right) \\ =-\sin \left(\frac{2\mathrm{\pi }}{3}\right)=-\frac{\sqrt{3}}{2} \end{gathered}$$

Vậy

$$\begin{gathered} \underset{x\to \infty }{\lim } \\ \left[\cos \left(\pi n\sqrt[3]{n^3+3n^2+n+1}\right)\right] \\ +\left[\sin \left(\mathrm{\pi n}\sqrt[3]{{\mathrm{n}}^3+3{\mathrm{n}}^2+\mathrm{n}+1}\right)\right] \\ =-\frac{1+\sqrt{3}}{2} \end{gathered}$$

Đáp án A