# How to evaluate limits of inverse trigonometric functions without L'Hospital and series expansion?

I am studying limits and how to evaluate them without using l'Hospital Rule or series expansion. Most of them aren't that hard, there are some common trick to do, but I have issues when I face limits of some not-so-common functions such as inverse trigonometric functions.

An example of such a function is this:

And also:

I have no idea how to even approach such a limit so I would be happy if you could, besides just solving these two limits explain some approaches to evaluating limits with inverse trigonometric functions in general.

$$\lim_{x\to0}\frac{\arctan(5x)}{\arctan(6x)} = \lim_{x\to0}\frac{\arctan(5x)}{\arctan(6x)}\cdot\frac{6x}{5x}\cdot\frac56 = \frac56\lim_{x\to0}\frac{\arctan(5x)}{5x}\lim_{x\to0}\frac{6x}{\arctan(6x)}=\frac56$$
Showing $$\lim_{u\to0}\frac{\arctan u}{u}=1$$
We know that $$\lim_{x\to0}\frac{\tan x }{ x } = \lim_{x\to 0} \frac{x}{\tan x} = 1$$ $$\text{ Now substitute }\tan x = u \implies x = \arctan u \text{ and as } x\to 0 , u = \tan x \to 0$$
• @Matthew Yeah, I've come across that one, but here it'd become $\arctan 3x$ – Ak. Mar 18 '20 at 15:06