$\lim_{n\to\infty} \arctan(\ln(n))$ $$\lim_{n\to\infty} \arctan(\ln(n)), n \geq 4$$ 
How do I properly say this equals $\pi/2$
Firstly, I know $\tan(x)$ is bounded at $$-\pi/2 \leq \tan(x) \leq \pi/2$$
I know $\ln(n)$ is a increasing function for $n \geq 4$. How do I show this?
 A: let $\ln(n) = u$
we have limit of $\arctan(u)$ as $e^u$ approaches $\infty$.
$\arctan(u)$ approaches $\pi/2$ as $e^u$ approaches $\infty$. 
A: *

*$\ln$ is a continuous function $[0,\infty] \to [-\infty, \infty]$.

*$\arctan$ is a continuous function $[-\infty,\infty] \to [-\pi/2, \pi/2]$.


Yes, I mean to include the endpoints; these should be understand as functions from and to the extended real numbers.
Consequently,
$$ \lim_{x \to \infty} \arctan(\ln(x)) = \arctan(\ln(\infty)) = \arctan(\infty) = \pi/2 $$

Of course, to use this argument one must know about the topology of the extended reals and that the relevant basic properties of limits hold. But with that knowledge, this is by far the easiest argument for the problem at hand.
This knowledge is almost taught in introductory calculus — it's what's going on 'behind the scenes' with the definition of limits involving $\pm \infty$.
A: To see how the limit is approached, remebering that $$\tan^{-1}(x)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac \pi 2$$ we then have $$\tan ^{-1}(\log (n))=\frac \pi 2-\tan ^{-1}\left(\frac{1}{\log (n)}\right)$$ Using now for small $x$  $$\tan ^{-1}(x)=x-\frac{x^3}{3}+O\left(x^4\right)$$ we then have $$\tan ^{-1}(\log (n))=\frac \pi 2-\frac{1}{\log (n)}+\frac{1}{3\log^3 (n)}+\cdots$$ which shows the limit and how it is appoached.
Consider $n=100$; the above trruncated expansion gives $\approx 1.35706$ while the exact value would be  $\approx 1.35697$.
A: You combine three simple results which are pretty standard in calculus:
1) $\lim_{n \to \infty}\log n = \infty$
2) $\lim_{x \to \infty}\arctan x = \dfrac{\pi}{2}$
3) Rule of Substitution for limits: If $\lim_{x \to a}f(x) = b$ and $f(x) \neq b$ as $x \to a$ and $\lim_{x \to b}g(x) = L$ then $\lim_{x \to a}g(f(x)) = L$. The rule holds even if $a, b$ are $\pm\infty$.
In the current question $a = b = \infty$ and $f(x) = \log x, g(x) = \arctan x$ and result follows from the above facts.
