Thread: Limit question of course!

How do I show that $\displaystyle \lim_{x\to\infty}\frac{\arcsin \left(\frac{1}{n^2}\right)}{\arctan \left(\frac{1}{n^2}\right)}=1$?

Thanks!

Originally Posted by polymerase

How do I show that $\displaystyle \lim_{x\to\infty}\frac{\arcsin \left(\frac{1}{n^2}\right)}{\arctan \left(\frac{1}{n^2}\right)}=1$?

Thanks!

$\displaystyle \lim_{n \to {\infty}}\frac{arcsin\bigg(\frac{1}{n^2}\bigg)}{ar ctan\bigg(\frac{1}{n^2}\bigg)}=\lim_{n \to {\infty}}\frac{arccsc(n^2)}{arcot(n^2)}=\frac{0}{0 }$ L'hopital's

Originally Posted by Mathstud28

$\displaystyle \lim_{n \to {\infty}}\frac{arcsin\bigg(\frac{1}{n^2}\bigg)}{ar ctan\bigg(\frac{1}{n^2}\bigg)}=\lim_{n \to {\infty}}\frac{arccsc(n^2)}{arcot(n^2)}=\frac{0}{0 }$ L'hopital's

you ge $\displaystyle \lim_{n \to {\infty}}\frac{1}{n^4\sqrt{1-\frac{1}{n^4}}}+\frac{n^4}{n^4\sqrt{1-\frac{1}{n^4}}}$

Originally Posted by polymerase

How do I show that $\displaystyle \lim_{x\to\infty}\frac{\arcsin \left(\frac{1}{{\color{red}x}^2}\right)}{\arctan \left(\frac{1}{{\color{red}x}^2}\right)}=1$?

Thanks!

I've made a correction (in red).

There's no emergency - no need for the hospital:

Let 1/x = t:

$\displaystyle \lim_{t \to 0}\frac{\arcsin \left(t^2\right)}{\arctan \left(t^2\right)}$

Substitute the appropriate Maclaurin series:

$\displaystyle = \lim_{t \to 0}\frac{t^2 + t^6/6 + ....}{t^2 - t^6/3 + ....}$

Divide the top and the bottom by t^2 and take the limit:

$\displaystyle = \lim_{t \to 0}\frac{1 + t^4/6 + ....}{1 - t^4/3 + ....} = 1$

Originally Posted by mr fantastic

I've made a correction (in red).

There's no emergency - no need for the hospital:

Let 1/x = t:

$\displaystyle \lim_{t \to 0}\frac{\arcsin \left(t^2\right)}{\arctan \left(t^2\right)}$

Substitute the appropriate Maclaurin series:

$\displaystyle = \lim_{t \to 0}\frac{t^2 + t^6/6 + ....}{t^2 - t^6/3 + ....}$

Divide the top and the bottom by t^2 and take the limit:

$\displaystyle = \lim_{t \to 0}\frac{1 + t^4/6 + ....}{1 - t^4/3 + ....} = 1$

You know what...I actually thought about doing it this way..but then i said...you know what would be much easier? L'hopital's!

Originally Posted by Mathstud28

You know what...I actually thought about doing it this way..but then i said...you know what would be much easier? L'hopital's!

Well ...... you say tomarto, I say tomayto