Results 1 to 5 of 5

Math Help - Limit of function

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    1

    Limit of function

    Hi everyone!

    First of all thank you to all of you who will answer.

    I am attending Calculus course at the university.

    I found this limit particularly hard for me to compute:

    Limit f x→inf = ?

    And f(x) = (3^x+7^x)^(1/x)

    I can only get to take it to an exponential form.

    I send it to an online limit solver and the answer was: 7.

    But I cannot past an indefinite inf^0

    Thank you very much.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by coltelino View Post
    Hi everyone!

    First of all thank you to all of you who will answer.

    I am attending Calculus course at the university.

    I found this limit particularly hard for me to compute:

    Limit f x→inf = ?

    And f(x) = (3^x+7^x)^(1/x)

    I can only get to take it to an exponential form.

    I send it to an online limit solver and the answer was: 7.

    But I cannot past an indefinite inf^0

    Thank you very much.
    use logs, or rewrite the function.

    Notice that (3^x + 7^x)^{1/x} = \exp (\ln [(3^x + 7^x)^{1/x}]) = \exp \left( \frac 1x \cdot \ln (3^x + 7^x) \right) = \exp \left( \frac {\ln (3^x + 7^x)}x\right)

    Now take the limit, apply L'Hopital's rule to finish up
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    As x approaches infinity,

    7^x becomes infinitely greater than 3^x

    Therefore the limit is (7^x)^{\frac{1}{x}}=7^{\frac{x}{x}}=7^1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,803
    Thanks
    1692
    Awards
    1
    Here is another way to see the problem.
    7 = \sqrt[n]{{7^n }} \leqslant \sqrt[n]{{3^n  + 7^n }} \leqslant 7\sqrt[n]{2} \to 7
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Very cool, Plato !
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: July 2nd 2011, 12:35 PM
  2. Finding limit of this function, using Limit rules
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 27th 2011, 01:12 PM
  3. Replies: 2
    Last Post: October 26th 2010, 10:23 AM
  4. Replies: 16
    Last Post: November 15th 2009, 04:18 PM
  5. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 3rd 2009, 05:05 PM

Search Tags


/mathhelpforum @mathhelpforum