## Calculus (3rd Edition)

$$\frac{\pi}{3}.$$
Using Theorem 1, we have $$\lim\limits_{n \to \infty}{\frac{n^3}{2n^3+1}}=\frac{1}{2}.$$ Now, since $\cos^{-1}{x}$ is continuous, then by Theorem 4, we have $$\lim_{n\to \infty}\cos^{-1}{\frac{n^3}{2n^3+1}}= \cos^{-1}{ \lim_{n\to \infty}\frac{n^3}{2n^3+1}}=\cos^{-1}\frac{1}{2}=\frac{\pi}{3}.$$