Results 1 to 5 of 5
Like Tree1Thanks
  • 1 Post By greg1313

Thread: Limit -- Variation on a problem

  1. #1
    Senior Member
    Joined
    Dec 2016
    From
    Earth
    Posts
    276
    Thanks
    145

    Limit -- Variation on a problem

    .


    $\displaystyle \displaystyle\lim_{n \to \infty}\bigg[n*e \ - \ n*(1 + \tfrac{1}{n})^n\bigg]$
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    20,249
    Thanks
    3366

    Re: Limit -- Variation on a problem

    First note that the discrete limit, $\displaystyle \lim_{n\to \infty} ne- n\left(n+ \frac{1}{n}\right)^n$ is the same as the continuous limit $\displaystyle \lim_{x\to\infty} xe- x\left(x+ \frac{1}{x}\right)^x$ as long as the latter limit exists. To do that, write it as $\displaystyle \lim_{x\to\infty}\frac{2- \left(x+ \frac{1}{x}\right)^x}{\frac{1}{x}}$, where both numerator go to 0, and use L'Hopital's rule.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,401
    Thanks
    3292
    Awards
    1

    Re: Limit -- Variation on a problem

    Quote Originally Posted by HallsofIvy View Post
    First note that the discrete limit, $\displaystyle \lim_{n\to \infty} ne- n\left(\color{red}n+ \frac{1}{n}\right)^n$ is the same as the continuous limit $\displaystyle \large\lim_{x\to\infty} xe- x\left(\color{red}x+ \frac{1}{x}\right)^x$ as long as the latter limit exists. To do that, write it as $\displaystyle \large\lim_{x\to\infty}\frac{\color{red}2- \left(\color{red}x+ \frac{1}{x}\right)^x}{\frac{1}{x}}$, where both numerator go to 0, and use L'Hopital's rule.
    Prof. Ivey, are there typo in the above?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    20,249
    Thanks
    3366

    Re: Limit -- Variation on a problem

    Not actually a typo- I just misread the problem!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    1,005
    Thanks
    510

    Re: Limit -- Variation on a problem

    we can use L'H˘pital's rule a couple of times or write

    $\displaystyle e-\left(1+\frac{1}{n}\right)^n=\sum _{k=2}^n \left(\frac{1}{k!}-\left(
    \begin{array}{c}
    n \\
    k
    \end{array}
    \right)\frac{1}{n^k}\right)+\sum _{k=n+1}^{\infty } \frac{1}{k!}$

    using the series for $e$ and the binomial theorem to expand $\left(1+\frac{1}{n}\right)^n$

    do some algebra then multiply by $n$ and take the limit
    Last edited by Idea; Mar 27th 2019 at 01:15 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inverse variation problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Aug 16th 2012, 03:01 AM
  2. variation problem
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Dec 9th 2010, 08:53 AM
  3. Variation problem
    Posted in the Algebra Forum
    Replies: 6
    Last Post: Dec 5th 2010, 02:05 PM
  4. Variation problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Dec 2nd 2010, 01:33 AM
  5. A variation problem
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: Apr 13th 2008, 11:23 PM

/mathhelpforum @mathhelpforum