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.
Best Answer
$$\lim_{x\to0}\frac{\arctan(5x)}{\arctan(6x)} = \lim_{x\to0}\frac{\arctan(5x)}{\arctan(6x)}\cdot\frac{6x}{5x}\cdot\frac56 = \frac56\lim_{x\to0}\frac{\arctan(5x)}{5x}\lim_{x\to0}\frac{6x}{\arctan(6x)}=\frac56$$
Showing $\lim_{u\to0}\frac{\arctan u}{u}=1$
We know that $$\lim_{x\to0}\frac{\tan x }{ x } = \lim_{x\to 0} \frac{x}{\tan x} = 1$$ $$\text{ Now substitute }\tan x = u \implies x = \arctan u \text{ and as } x\to 0 , u = \tan x \to 0 $$