A problem in using theorem for finding limit

Question

Let $a_n = \arctan ( \ln n )$ , find $\lim_{ n \to \infty} a_n$ if any . I think we can't apply the theorem (i.e $\lim_{ n \to \infty} \arctan ( \ln n ) = \arctan (\lim_{ n \to \infty} \ln n) = \pi /2 )$ . The condition of theorem is $a_n \to L$ but here $a_n$ diverges to infinity . Although , we know that $\pi / 2$ is the right answer since $\lim_{x \to \infty} \arctan x = \pi /2 $ and instead of $a_n$ we can put any other sequence that diverges to infinity .

Answer 1

Since $arctan(x)=\pi/2-arctan(1/x)$ and $arctan(x) \to 0$ as $x \to 0$, if $f(n) \to \infty$ then $arctan(f(n))=\pi/2-arctan(1/f(n)) \to \pi/2$.

Answer 2

I believe you are correct and that you cannot use that theorem (which would require arctan to be continuous at infinity. You have touched on another way to do this in the end of your comment; In particular, using the sequential characterization of limits.

Since $\lim_{x \to \infty} \arctan(x) = \pi/2$, we know that $\arctan(x_n)\to \pi/2$ for any $x_n \to \infty$. Since $\ln(n) \to \infty$ we are done.

Answer 3

Since $b_n=\log n \to +\infty$

$$\lim_{n\to +\infty} a_n = \lim_{n\to +\infty} \arctan ( \ln n )= \lim_{b_n\to +\infty} \arctan ( b_n )=\frac{\pi}{2}$$