Determine the following limits

Question

$$\lim_{n\to ∞}\arctan(n!)$$
$$\lim_{n\to\infty}\frac{\sin(n\pi^2)\ln(1+n)}{n}$$

I don't have a solution for these limits. If you could please give a hint to me, it would be appreciated.

Answer 1

Note that by definition

$$\arctan n!\to\frac{\pi}2$$

and $$\frac{\sin(n\pi^2)\ln(1+n)}{n}\to 0$$

indeed by squeeze theorem

$$0\le\left|\frac{\sin(n\pi^2)\ln(1+n)}{n}\right|\le \frac{\ln(1+n)}{n}\to 0$$

Answer 2

As you know, $\arctan(n) \to \frac{\pi}{2}, n \to \infty$ hence your first limit is equal to $\frac{\pi}{2}$ As for your second limit, let's consider $\frac{\ln (1 + n)}{n}$ this sequence $\to 0$ because $\ln$ increases much slower than a linear function. So, your second limit is a product of the bounded function $\sin$ and the infinitesimally small function, which $\to 0$. Thus, your second limit is equal to $0$.