Results 1 to 3 of 3

Math Help - Compute the following limit (Use L'H˘pital's rule)

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    50

    Compute the following limit (Use L'H˘pital's rule)

    Compute the following limit (Use L'H˘pital's rule):

    [lncos(a/x)]/x^(-2) as x tends to infinity.
    Last edited by mr fantastic; May 6th 2010 at 06:05 PM. Reason: Copied from title into main body of post.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by feyomi View Post
    [lncos(a/x)]/x^(-2) as x tends to infinity.
    Hi feyomi,

    \frac{\frac{d}{dx}\ ln\left[cos\left(\frac{a}{x}\right)\right]}{\frac{d}{dx}x^{-2}}=\frac{\frac{d}{dx}lnu}{-2x^{-3}}=\frac{\frac{du}{dx}\frac{d}{du}lnu}{-2x^{-3}}=\frac{\frac{d}{dx}\left[cos\left(\frac{a}{x}\right)\right]\frac{1}{u}}{-2x^{-3}}

    =\frac{\left[\frac{dv}{dx}\frac{d}{dv}cosv\right]\frac{1}{u}}{-2x^{-3}}=\frac{\frac{-a}{x^2}\left[-sin\left(\frac{a}{x}\right)\right]}{-2x^{-3}cos\left(\frac{a}{x}\right)}=-\frac{asin\left(\frac{a}{x}\right)}{2x^2x^{-3}cos\left(\frac{a}{x}\right)}

    =-\frac{a^2sin\left(\frac{a}{x}\right)}{2\left(\frac  {a}{x}\right)}\ \frac{1}{cos\left(\frac{a}{x}\right)}

    You can now evaluate the limit, since as x goes to infinity

    \frac{a}{x}\ \rightarrow\ 0

    \lim_{u\rightarrow\ 0}\frac{sinu}{u}=1

    cos(0)=1

    hence the limit is -\frac{a^2}{2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,969
    Thanks
    1011
    Quote Originally Posted by feyomi View Post
    [lncos(a/x)]/x^(-2) as x tends to infinity.
    This is a bit hard to read...

    Is it

    \lim_{x \to \infty}{\frac{\ln{\left[\cos{\left(\frac{a}{x}\right)}\right]}}{x^{-2}}}?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limit using L'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 6th 2011, 05:28 AM
  2. [SOLVED] Tricky limit without using l'Hopital's rule
    Posted in the Calculus Forum
    Replies: 9
    Last Post: December 15th 2010, 02:21 PM
  3. Replies: 1
    Last Post: April 24th 2010, 02:45 PM
  4. Replies: 1
    Last Post: April 24th 2010, 01:40 PM
  5. Limit w. conjugate/ L'Hopital's Rule
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: September 17th 2008, 11:08 AM

Search Tags


/mathhelpforum @mathhelpforum