Results 1 to 6 of 6

Math Help - Limit question of course!

  1. #1
    Senior Member polymerase's Avatar
    Joined
    May 2007
    From
    Sydney
    Posts
    267

    Limit question of course!

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by polymerase View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Mathstud28 View Post
    \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}}}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by polymerase View Post
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by mr fantastic View Post
    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!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Mathstud28 View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: November 7th 2011, 03:27 PM
  2. Replies: 1
    Last Post: August 8th 2010, 11:29 AM
  3. Limit question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 21st 2009, 05:04 AM
  4. Limit question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 26th 2009, 11:00 AM
  5. limit question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 6th 2006, 10:30 AM

Search Tags


/mathhelpforum @mathhelpforum