Results 1 to 4 of 4

Thread: Squeeze Theorem

  1. #1
    Member
    Joined
    Dec 2007
    Posts
    137

    Red face Squeeze Theorem

    How would I use the squeeze theorem to evaluate for:
    $\displaystyle lim_{n\to\infty}\{\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2 + n}} \} $
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,781
    Thanks
    2823
    Awards
    1
    $\displaystyle \frac{n}
    {{\sqrt {n^2 + n} }} \leqslant \sum\limits_{k = 1}^n {\frac{1}
    {{\sqrt {n^2 + k} }}} \leqslant \frac{n}
    {{\sqrt {n^2 + 1} }}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2007
    Posts
    137
    So after that step I would solve for the limits of those outer functions,
    how would I start for:
    $\displaystyle lim_{n\to\infty}\frac{n}{\sqrt{n^2+n}}$
    I was going to factor out the highest power of "n" but don't know how to apply that in the denominator since it's under a square root sign.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2
    Quote Originally Posted by xxlvh View Post
    So after that step I would solve for the limits of those outer functions,
    how would I start for:
    $\displaystyle lim_{n\to\infty}\frac{n}{\sqrt{n^2+n}}$
    I was going to factor out the highest power of "n" but don't know how to apply that in the denominator since it's under a square root sign.
    Plato already completed the proof, but there is somethings missing with you.
    I hope the following will fill it.
    $\displaystyle \lim_{n\to\infty}\frac{n}{\sqrt{n^{2}+n}}=\lim_{n\ to\infty}\frac{1}{\frac{\sqrt{n^{2}+n}}{n}}$
    $\displaystyle =\lim_{n\to\infty}\frac{1}{\sqrt{\frac{n^{2}+n}{n^ {2}}}}$ (as $\displaystyle n\to\infty$, we may take $\displaystyle n>0$)
    $\displaystyle =\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{1}{n}}}$
    $\displaystyle =\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{1}{n}}}=1$ (since $\displaystyle 1/\sqrt{1+\cdot}$ is continuous in a neighbourhood of $\displaystyle 0$)
    $\displaystyle =\frac{1}{\sqrt{1+\lim\limits_{n\to\infty}\frac{1} {n}}}=\frac{1}{\sqrt{1+0}}=1$
    I guess you can show now the limit on the right-hand side...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Squeeze Theorem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Sep 8th 2010, 04:00 PM
  2. Squeeze theorem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 8th 2009, 07:43 PM
  3. Need help with example of Squeeze Theorem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 26th 2009, 04:03 AM
  4. squeeze theorem
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Apr 22nd 2009, 02:49 AM
  5. help with squeeze theorem.
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: Jan 8th 2008, 03:59 PM

Search Tags


/mathhelpforum @mathhelpforum