Results 1 to 7 of 7
Like Tree2Thanks
  • 1 Post By MarkFL
  • 1 Post By MarkFL

Thread: limits ( the indeterminate forms)

  1. #1
    Junior Member romeobernard's Avatar
    Joined
    Oct 2018
    From
    PH
    Posts
    33
    Thanks
    3

    limits ( the indeterminate forms)

    Hi guys, anyone can help me about this topic.

    Thanks a lot.

    Sent from my SM-J730G using Tapatalk
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,239
    Thanks
    944

    Re: limits ( the indeterminate forms)

    Is it:

    $\displaystyle \lim_{x\to\infty}\left(\left(1+x^2e^x\right) ^{\Large\frac{1}{x}}\right)$ ?
    Thanks from romeobernard
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member romeobernard's Avatar
    Joined
    Oct 2018
    From
    PH
    Posts
    33
    Thanks
    3

    Re: limits ( the indeterminate forms)

    Quote Originally Posted by MarkFL View Post
    Is it:

    $\displaystyle \lim_{x\to\infty}\left(\left(1+x^2e^x\right) ^{\Large\frac{1}{x}}\right)$ ?
    yes thats exactly the problem.

    Sent from my SM-J730G using Tapatalk
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,239
    Thanks
    944

    Re: limits ( the indeterminate forms)

    Quote Originally Posted by romeobernard View Post
    yes thats exactly the problem.
    Okay, what I would do is set:

    $\displaystyle \lim_{x\to\infty}\left(\left(1+x^2e^x\right) ^{\Large\frac{1}{x}}\right)=L$

    And take the natural log of both sides:

    $\displaystyle \ln\left(\lim_{x \to\infty}\left( \left(1+x^2e^x \right) ^{\Large\frac{1}{x}}\right)\right) =\ln(L)$

    Now, because log functions are one-to-one, we can bring it inside the limit:

    $\displaystyle \lim_{x\to\infty} \left(\ln\left( \left(1+x^2e^x \right) ^{\Large \frac{1}{x}}\right) \right) =\ln(L)$

    Now, applying a log property, we may write:

    $\displaystyle \lim_{x\to\infty} \left(\frac{\ln \left(1+x^2e^x\right)}{x} \right) =\ln(L)$

    Now, we have the indeterminate form $\displaystyle \frac{\infty}{\infty}$ on the LHS and can apply L'H˘pital's Rule...what do you get?
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member romeobernard's Avatar
    Joined
    Oct 2018
    From
    PH
    Posts
    33
    Thanks
    3

    Re: limits ( the indeterminate forms)

    Quote Originally Posted by MarkFL View Post
    Okay, what I would do is set:

    $\displaystyle \lim_{x\to\infty}\left(\left(1+x^2e^x\right) ^{\Large\frac{1}{x}}\right)=L$

    And take the natural log of both sides:

    $\displaystyle \ln\left(\lim_{x \to\infty}\left( \left(1+x^2e^x \right) ^{\Large\frac{1}{x}}\right)\right) =\ln(L)$

    Now, because log functions are one-to-one, we can bring it inside the limit:

    $\displaystyle \lim_{x\to\infty} \left(\ln\left( \left(1+x^2e^x \right) ^{\Large \frac{1}{x}}\right) \right) =\ln(L)$

    Now, applying a log property, we may write:

    $\displaystyle \lim_{x\to\infty} \left(\frac{\ln \left(1+x^2e^x\right)}{x} \right) =\ln(L)$

    Now, we have the indeterminate form $\displaystyle \frac{\infty}{\infty}$ on the LHS and can apply L'H˘pital's Rule...what do you get?
    thanks wait i will rewrite this equation and see if i can get this.

    Sent from my SM-J730G using Tapatalk
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,239
    Thanks
    944

    Re: limits ( the indeterminate forms)

    To follow up, using L'H˘pital's Rule, we may write:

    $\displaystyle \lim_{x\to\infty} \left(\frac{\dfrac{2xe^x+x^2e^x}{1+x^2e^x}}{1} \right) =\ln(L)$

    Or:

    $\displaystyle \lim_{x\to\infty} \left(\frac{x^2+2x}{e^{-x}+x^2} \right) =\ln(L)$

    Applying L'H˘pital's Rule again, we get:

    $\displaystyle \lim_{x\to\infty} \left(\frac{2x+2}{2x-e^{-x}} \right) =\ln(L)$

    Or:

    $\displaystyle \lim_{x\to\infty} \left(\frac{2+ \dfrac{2} {x}}{2- \dfrac{1}{xe^{x}}} \right) =\ln(L)$

    Hence:

    $\displaystyle 1=\ln(L)\implies L=e$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2016
    From
    Earth
    Posts
    227
    Thanks
    109

    Re: limits ( the indeterminate forms)

    You can avoid L'H˘pital's Rule if you would show that


    $\displaystyle \displaystyle\lim_{x \to \infty}(e^x)^{1/x} \ \le \ \displaystyle\lim_{x \to \infty}(1 + x^2e^x)^{1/x} \ \le \ \displaystyle\lim_{x \to \infty}(e^\sqrt{x}e^x)^{1/x} \ \ \ \ $**


    $\displaystyle e \ \le \ \displaystyle\lim_{x \to \infty}(1 + x^2e^x)^{1/x} \ \le \ \displaystyle\lim_{x \to \infty}(e^{x + \sqrt{x}})^{1/x}$


    $\displaystyle e \ \le \ \displaystyle\lim_{x \to \infty}(1 + x^2e^x)^{1/x} \ \le \ \displaystyle\lim_{x \to \infty}(e^{1 + \tfrac{1}{\sqrt{x}}}) $


    $\displaystyle e \ \le \ \displaystyle\lim_{x \to \infty}(1 + x^2e^x)^{1/x} \ \le \ e $




    ** You would want to make sure that you can show that $\displaystyle \ \ \displaystyle\lim_{x \to \infty}\bigg(\dfrac{e^\sqrt{x}e^x}{1 + x^2e^x}\bigg) > 1$.
    Last edited by greg1313; Dec 4th 2018 at 03:28 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Abstraction and Indeterminate Forms of Limits
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: Jul 13th 2012, 01:28 PM
  2. please help with limits of indeterminate forms
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Mar 19th 2009, 08:18 PM
  3. Limits & Indeterminate Forms
    Posted in the Calculus Forum
    Replies: 10
    Last Post: Oct 25th 2007, 02:47 PM
  4. LImits at infinity help: Indeterminate forms
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Mar 11th 2007, 10:02 AM
  5. Indeterminate Forms: Limits infinite... help
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 4th 2007, 05:06 AM

/mathhelpforum @mathhelpforum