Results 1 to 3 of 3

Math Help - [SOLVED] Help finding limits that approach ∞.

  1. #1
    Member ilikedmath's Avatar
    Joined
    Sep 2008
    Posts
    98

    Exclamation [SOLVED] Help finding limits that approach ∞.

    (1) ==

    I rationalized the expression by multiplying the numerator and denominator by the conjugate of the denominator, but then I got stuck and didn't know where to go from there. It seems that my rationalizing just made finding the limit more complicated. Would I then divide everything by what dominates the denominator? So "n" dominates the denominator so if I divide everything by n, I get: . I know that (1/n) goes to 0, but how can I get rid of the '3n'?

    (2) =. Then I divided everything by 7^n since it dominates the denominator. I get: .
    I can rewrite (4^n)/(7^n) as (4/7)^n. And since (4/7)^n goes to 0, the expression further simplifies to:
    . But then if (4/7)^n goes to 0, I get 11/0 which is undefined. So I'm wrong.

    Do I need to use the squeeze theorem for these? I haven't had calculus in two years, so I am very rusty. This is for a intro to real analysis class.

    Any help, tips, corrections, and/or suggestions are greatly appreciated.
    Thank you for your time!

    - Stumped Student
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, ilikedmath!

    1)\;\;\lim_{n\to\infty}\frac{\sqrt{3n-1}}{\sqrt{n}+2}
    Divide top and bottom by \sqrt{n}

    \lim_{n\to\infty}\left(\frac{\dfrac{\sqrt{3n-1}}{\sqrt{n}}} {\dfrac{\sqrt{n}}{\sqrt{n}} + \dfrac{2}{\sqrt{n}}}\right) \;=\;\lim_{n\to\infty}\left( \frac{\sqrt{\dfrac{3n}{n} - \dfrac{1}{n}}} {1 + \dfrac{2}{\sqrt{n}}}\right) . =\;\lim_{n\to\infty}\left(\frac{\sqrt{3 - \dfrac{1}{n}}}{1 + \dfrac{2}{\sqrt{n}}}\right) \;=\;\frac{\sqrt{3-0}}{1+0} \;=\;\sqrt{3}




    2)\;\;\lim_{n\to\infty}\frac{4^{n+1} + 7^{n+1}}{4^n + 7^n}
    Divide top and bottom by 7^n

    \lim_{n\to\infty}\left[ \frac{\dfrac{4^{n+1}}{7^n} + \dfrac{7^{n+1}}{7^n}} {\dfrac{4^n}{7^n} + \dfrac{7^n}{7^n}}\right] \;=\;\lim_{n\to\infty}\left[ \frac{4\left(\dfrac{4}{7}\right)^n + 7} {\left(\dfrac{4}{7}\right)^n + 1}\right] . = \;\frac{4\!\cdot\!0 + 7}{0 + 1} \;=\;7

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member ilikedmath's Avatar
    Joined
    Sep 2008
    Posts
    98
    Quote Originally Posted by Soroban View Post
    Hello, ilikedmath!
    <-- me laughing at myself for making it a lot more complicated than it was.

    Thank you very much!!! Very clear and helpful answer you gave, A+++!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Systematic approach for multi variable limits?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: December 18th 2011, 05:31 PM
  2. Replies: 2
    Last Post: December 18th 2011, 05:13 PM
  3. Finding the limit for x->∞
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 10th 2009, 08:49 PM
  4. [SOLVED] finding limits
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 6th 2009, 04:59 PM
  5. Replies: 5
    Last Post: March 23rd 2009, 07:21 AM

Search Tags


/mathhelpforum @mathhelpforum