Results 1 to 3 of 3

Thread: L'Hopital Limit

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    11

    L'Hopital Limit

    $\displaystyle \lim_{x\to\infty}x{[(1+\frac{1}{x})^x -e]}$ I know its indeterminate form $\displaystyle \infty*0$ and so i rewrite it as
    $\displaystyle \lim_{x\to\infty}\frac{(1+\frac{1}{x})^x -e}{\frac{1}{x}}$. Now it's $\displaystyle \frac{0}{0}$ indeterminate form. Then I get stuck.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by xxsteelxx View Post
    $\displaystyle \lim_{x\to\infty}x{[(1+\frac{1}{x})^x -e]}$ I know its indeterminate form $\displaystyle \infty*0$ and so i rewrite it as
    $\displaystyle \lim_{x\to\infty}\frac{(1+\frac{1}{x})^x -e}{\frac{1}{x}}$. Now it's $\displaystyle \frac{0}{0}$ indeterminate form. Then I get stuck.

    This is really a tough limit: first, and in order to derivate more easily, put

    $\displaystyle \displaystyle {\left(1 + \frac{1}{x}\right)^x= e^{x\ln \left(1 + \frac{1}{x}\right)}}$

    so that $\displaystyle \left(\left(1 + \frac{1}{x}\right)^x\right)'$$\displaystyle =\left(1 + \frac{1}{x}\right)^x \left[\ln \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right]$

    Applying now L'Hospital to your limit we get $\displaystyle -x^2\left(1 + \frac{1}{x}\right)^x \left[\ln \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right]$.

    Let apart the $\displaystyle -\left(1 +\frac{1}{x}\right)^x$ term, whose limit is $\displaystyle -e$ , and focus on the rest:

    $\displaystyle \frac{\ln \left(1+\frac{1}{x}\right)-\frac{1}{x+1}}{\frac{1}{x^2}}$

    Apply L'Hospital one again here and get $\displaystyle \frac{-\frac{1}{x(x+1)}+\frac{1}{(x+1)^2}}{-\frac{2}{x^3}}$ $\displaystyle =\frac{1}{2}\left[\frac{x^2}{x+1}-\frac{x^3}{(x+1)^2}\right]$$\displaystyle =\frac{1}{2}\frac{x^2}{(x+1)^2} \xrightarrow [x \to \infty] {} \frac{1}{2}$

    Now don't forget the term you left out before and the limit finally is (**Pant, pant!!**) : $\displaystyle -\frac{e}{2}$

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    11
    Thank you very much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. L'Hopital's limit
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Apr 27th 2011, 05:52 PM
  2. Limit using L'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Apr 6th 2011, 05:28 AM
  3. Limit using L'H˘pital
    Posted in the Calculus Forum
    Replies: 11
    Last Post: May 25th 2010, 09:52 AM
  4. limit/l'hopital
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Nov 14th 2009, 05:51 PM
  5. limit using l'hopital
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Dec 24th 2007, 09:50 AM

Search Tags


/mathhelpforum @mathhelpforum