Limit question of course!

**polymerase** · April 19th 2008, 07:46 PM

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

Thanks!

**Mathstud28** · April 19th 2008, 07:51 PM

Originally Posted by polymerase

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

Thanks!

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

**Mathstud28** · April 19th 2008, 07:56 PM

Originally Posted by Mathstud28

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

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

**mr fantastic** · April 19th 2008, 08:46 PM

Originally Posted by polymerase

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

Thanks!

I've made a correction (in red).

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

Let 1/x = t:

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

Substitute the appropriate Maclaurin series:

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

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

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

**Mathstud28** · April 19th 2008, 08:56 PM

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:

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

Substitute the appropriate Maclaurin series:

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

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

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

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!

**mr fantastic** · April 19th 2008, 09:06 PM

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

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