3
$\begingroup$

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:

enter image description here

And also:

enter image description here

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.

Improve this question
$\endgroup$
3
$\begingroup$

$$\lim_{x\to0}\frac{\arctan(5x)}{\arctan(6x)} = \lim_{x\to0}\frac{\arctan(5x)}{\arctan(6x)}\cdot\frac{6x}{5x}\cdot\frac56 = \frac56\lim_{x\to0}\frac{\arctan(5x)}{5x}\lim_{x\to0}\frac{6x}{\arctan(6x)}=\frac56$$

Showing $\lim_{u\to0}\frac{\arctan u}{u}=1$

We know that $$\lim_{x\to0}\frac{\tan x }{ x } = \lim_{x\to 0} \frac{x}{\tan x} = 1$$ $$\text{ Now substitute }\tan x = u \implies x = \arctan u \text{ and as } x\to 0 , u = \tan x \to 0 $$

Improve this answer
$\endgroup$
  • 1
    $\begingroup$ Thank you, but how do I solve the second example? $\endgroup$ – Matthew Mar 18 '20 at 14:51
  • 1
    $\begingroup$ @Matthew I'm working on that one. There must be some trick to evaluate it w/o L' Hopital or expansion. $\endgroup$ – Ak. Mar 18 '20 at 14:53
  • 1
    $\begingroup$ It didn't help me, but maybe you will find some use in this limit that looks very similar to the one I'm asking about: socratic.org/questions/…. $\endgroup$ – Matthew Mar 18 '20 at 15:02
  • 1
    $\begingroup$ @Matthew Yeah, I've come across that one, but here it'd become $\arctan 3x$ $\endgroup$ – Ak. Mar 18 '20 at 15:06
  • 1
    $\begingroup$ Someone commented the following link: math.stackexchange.com/questions/387333/…, where this exact limit is calculated. Thanks again. $\endgroup$ – Matthew Mar 18 '20 at 16:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.