Results 1 to 4 of 4

Math Help - How could I solve this limit with Taylor series?

  1. #1
    Newbie dttah's Avatar
    Joined
    Nov 2011
    Posts
    18

    How could I solve this limit with Taylor series?

    Hello everyone.

    I am not sure how to solve this with Taylor/Mc-Laurin series:

    \frac{log(e^x-x-x^2)}{logsinx-logx}

    I have no problems at the numerator, I mean, I could do Mc-Laurin series for e^x so then I'd have a "1" which could help out in the following Mc-Laurin development:

    log(1+x).

    I can't help myself at the denominator though. Am I obbliged to do Taylor series with center in 1?
    And if I do, should I change the others too?
    I was thinking even about applying log's properties, but that doesn't bring me too far.
    Any suggestions?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7

    Re: How could I solve this limit with Taylor series?

    Quote Originally Posted by dttah View Post
    Hello everyone.

    I am not sure how to solve this with Taylor/Mc-Laurin series:

    \frac{\log(e^x-x-x^2)}{log\sin x-\log x}

    I have no problems at the numerator, I mean, I could do Mc-Laurin series for e^x so then I'd have a "1" which could help out in the following Mc-Laurin development:

    log(1+x).

    I can't help myself at the denominator though. Am I obliged to do Taylor series with center in 1?
    And if I do, should I change the others too?
    I was thinking even about applying log's properties, but that doesn't bring me too far.
    Any suggestions?
    For the denominator, follow your own suggestion and use properties of logs:

    \begin{aligned}\log\sin x-\log x = \log\bigl(\tfrac{\sin x}{x}\bigr) &= \log\bigl(\tfrac{x-x^3/3! + x^5/5!-\ldots}x\bigr) \\ &= \log\bigl(1-(x^2/3!-x^4/5!+\ldots)\bigr) = \ldots .\end{aligned}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,547
    Thanks
    1418

    Re: How could I solve this limit with Taylor series?

    Quote Originally Posted by dttah View Post
    Hello everyone.

    I am not sure how to solve this with Taylor/Mc-Laurin series:

    \frac{log(e^x-x-x^2)}{logsinx-logx}

    I have no problems at the numerator, I mean, I could do Mc-Laurin series for e^x so then I'd have a "1" which could help out in the following Mc-Laurin development:

    log(1+x).

    I can't help myself at the denominator though. Am I obbliged to do Taylor series with center in 1?
    And if I do, should I change the others too?
    I was thinking even about applying log's properties, but that doesn't bring me too far.
    Any suggestions?
    What value are you making x approach?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie dttah's Avatar
    Joined
    Nov 2011
    Posts
    18

    Re: How could I solve this limit with Taylor series?

    Oh, righty, since I used the log's properties, I just could have done the senx/x taylor series. Thanks a lot!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: May 6th 2010, 12:49 PM
  2. Taylor Series on a limit
    Posted in the Calculus Forum
    Replies: 7
    Last Post: April 9th 2009, 07:10 AM
  3. Find limit using taylor series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 9th 2009, 06:49 AM
  4. solving limit using taylor series..
    Posted in the Calculus Forum
    Replies: 7
    Last Post: March 10th 2009, 07:33 AM
  5. Replies: 9
    Last Post: April 3rd 2008, 05:50 PM

Search Tags


/mathhelpforum @mathhelpforum