Results 1 to 6 of 6

Math Help - Limits

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    221

    Limits

    Calculate the following limits:

    1.) \lim_{x \to \infty} \frac{\ln{x}}{x^{\frac{2}{3}}}

    I know the answer is 0. Can I simply use L'Hopitals?

    So, \frac{\frac{1}{x}}{\frac{2}{3}x^{\frac{-1}{3}}}? Something like that?

    And then,

    2.) Does \int_{1}^{\infty} \frac{e^{-x}}{\sqrt{x}}\, dx converge? Justify your answer.

    No idea how to justify, but my calculator gives .2788 as an answer so it def does converge.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Ideasman View Post
    Calculate the following limits:

    1.) \lim_{x \to \infty} \frac{\ln{x}}{x^{\frac{2}{3}}}

    I know the answer is 0. Can I simply use L'Hopitals?

    So, \frac{\frac{1}{x}}{\frac{2}{3}x^{\frac{-1}{3}}}? Something like that?

    And then,

    2.) Does \int_{1}^{\infty} \frac{e^{-x}}{\sqrt{x}}\, dx converge? Justify your answer.

    No idea how to justify, but my calculator gives .2788 as an answer so it def does converge.
    Yes you can use l'hopitals
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    For the first one, i guess it's correct.

    For the second one : if the function in the integral is continuous and the limits at the extremities of the integral are finite, then the integral converges.

    In [1;+\infty[, \frac{e^{-x}}{\sqrt{x}} is continuous.
    In 1, the function has a finite value.

    Now let's study the limit at +\infty

    \frac{e^{-x}}{\sqrt{x}}=\frac{1}{e^x \sqrt{x}}

    The limit is clearly 0.

    So the integral converges.




    I hope i didn't miss any point :s
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Ideasman View Post
    Calculate the following limits:

    1.) \lim_{x \to \infty} \frac{\ln{x}}{x^{\frac{2}{3}}}

    I know the answer is 0. Can I simply use L'Hopitals?

    So, \frac{\frac{1}{x}}{\frac{2}{3}x^{\frac{-1}{3}}}? Something like that?

    And then,

    2.) Does \int_{1}^{\infty} \frac{e^{-x}}{\sqrt{x}}\, dx converge? Justify your answer.

    No idea how to justify, but my calculator gives .2788 as an answer so it def does converge.
    since imputting ∞ into the top and bottom gives \frac{\infty}{\infty}...you can use L'hopital's rule...so \lim_{x \to {\infty}}\frac{\ln(x)}{x^{\frac{2}{3}}}=\lim_{x \to {\infty}}\frac{\frac{1}{x}}{\frac{2x^{\frac{1}{3}}  }{3}}=\lim_{x \to {\infty}}\frac{2}{3x^{\frac{4}{3}}}=0
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,659
    Thanks
    1616
    Awards
    1
    Quote Originally Posted by Ideasman View Post
    Calculate the following limits:
    1.) \lim_{x \to \infty} \frac{\ln{x}}{x^{\frac{2}{3}}}
    I know the answer is 0. Can I simply use L'Hopitals?
    You do not need L'Hopitalsí rule here. It is a lazy persons way out.
    I am among those who would like to see it dropped from textbooks in basic calculus.
    Note \ln (x) = 3\ln \left( {x^{\frac{1}{3}} } \right) \le 3\left( {x^{\frac{1}{3}} } \right).

    Therefore, \frac{{\ln (x)}}{{x^{\frac{2}{3}} }} \le \frac{{3\left( {x^{\frac{1}{3}} } \right)}}{{x^{\frac{2}{3}} }} = \frac{3}{{x^{\frac{1}{3}} }} now the limit is so clear!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    So plato

    Quote Originally Posted by Plato View Post
    You do not need L'Hopitalsí rule here. It is a lazy persons way out.
    I am among those who would like to see it dropped from textbooks in basic calculus.
    Note \ln (x) = 3\ln \left( {x^{\frac{1}{3}} } \right) \le 3\left( {x^{\frac{1}{3}} } \right).

    Therefore, \frac{{\ln (x)}}{{x^{\frac{2}{3}} }} \le \frac{{3\left( {x^{\frac{1}{3}} } \right)}}{{x^{\frac{2}{3}} }} = \frac{3}{{x^{\frac{1}{3}} }} now the limit is so clear!
    Is this sort of like how we can't use L'hopital's rule on \lim_{n \to {\infty}}\frac{n^2}{n!} we know that \forall{x}\in[4,\infty],x^2<x! so we know the limit is 0?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Using limits to find other limits
    Posted in the Calculus Forum
    Replies: 7
    Last Post: September 18th 2009, 05:34 PM
  2. Function limits and sequence limits
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: April 26th 2009, 01:45 PM
  3. HELP on LIMITS
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 23rd 2008, 11:17 PM
  4. Limits
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 21st 2008, 10:52 PM
  5. [SOLVED] [SOLVED] Limits. LIMITS!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 25th 2008, 10:41 PM

Search Tags


/mathhelpforum @mathhelpforum