Results 1 to 12 of 12

Math Help - why didn't i solve the lim well?

  1. #1
    Junior Member
    Joined
    Mar 2013
    From
    israel
    Posts
    70

    why didn't i solve the lim well?

     \lim_{x\rightarrow-\infty}(\sqrt{x^{2}+x}-x) = ? . this is how i solve this, but i know it's not true.. why?  <br />
(\sqrt{x^{2}+x}-x)=\frac{(\sqrt{x^{2}+x}-x)\cdot(\sqrt{x^{2}+x}+x)}{(\sqrt{x^{2}+x}+x)}= \frac{x}{\sqrt{x^{2}+x}+x}
    \frac{x}{x+x\cdot\sqrt{1+\frac{1}{x}}}=\frac{1}{1+  \sqrt{1+\frac{1}{x}}} so,  \lim_{x\rightarrow-\infty}f(x)=\frac{1}{2}<br />
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jul 2012
    From
    INDIA
    Posts
    833
    Thanks
    209

    Re: why didn't i solve the lim well?

    I find the approach absolutely correct.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: why didn't i solve the lim well?

    I don't see where you went wrong myself, and your result agrees with mine.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2013
    From
    israel
    Posts
    70

    Re: why didn't i solve the lim well?

    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: why didn't i solve the lim well?

    Yes, and as x\to\infty the limit is 1/2.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Mar 2013
    From
    israel
    Posts
    70

    Re: why didn't i solve the lim well?

    Quote Originally Posted by MarkFL View Post
    Yes, and as x\to\infty the limit is 1/2.
    and what about the limit as x\rightarrow-\infty , which it's the case we're dealing with?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: why didn't i solve the lim well?

    I missed the negative sign...in that case I would rewrite the limit as:

    \lim_{x\to\infty}\left(\sqrt{x^2-x}+x \right)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Mar 2013
    From
    israel
    Posts
    70

    Re: why didn't i solve the lim well?

    didn't understand you.. what in this case?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: why didn't i solve the lim well?

    In the case of x\to-\infty I substituted x with -x...
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Apr 2012
    From
    Erewhon
    Posts
    200
    Thanks
    131

    Re: why didn't i solve the lim well?

    Quote Originally Posted by orir View Post
     \lim_{x\rightarrow-\infty}(\sqrt{x^{2}+x}-x) = ? . this is how i solve this, but i know it's not true.. why?  (\sqrt{x^{2}+x}-x)=\frac{(\sqrt{x^{2}+x}-x)\cdot(\sqrt{x^{2}+x}+x)}{(\sqrt{x^{2}+x}+x)}= \frac{x}{\sqrt{x^{2}+x}+x}
    \frac{x}{x+x\cdot\sqrt{1+\frac{1}{x}}}=\frac{1}{1+  \sqrt{1+\frac{1}{x}}} so,  \lim_{x\rightarrow-\infty}f(x)=\frac{1}{2}
    Because at the begining of the last line you should have had:

    \frac{x}{x+|x|\cdot\sqrt{1+\frac{1}{x}}}

    rather than:

    \frac{x}{x+x\cdot\sqrt{1+\frac{1}{x}}}

    If you want to try going down that route.

    .
    Last edited by zzephod; April 26th 2013 at 03:54 AM.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Mar 2013
    From
    israel
    Posts
    70

    Re: why didn't i solve the lim well?

    ok.. so i still don't get how it should be solved
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,908
    Thanks
    1759
    Awards
    1

    Re: why didn't i solve the lim well?

    Quote Originally Posted by orir View Post
    i still don't get how it should be solved

    Please note the problem is  \lim_{x\rightarrow{\color{red}-}\infty}(\sqrt{x^{2}+x}-x) = ?

    In effect it is \infty{\color{blue}+}\infty.

    To see that let x=-10^6 then we have \sqrt{10^{12}-10^6}+10^6=10^3\sqrt{10^2-1}+10^6, which is large.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: October 23rd 2012, 12:00 PM
  2. Replies: 2
    Last Post: November 8th 2009, 03:18 PM
  3. Several questions I didn't get
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 6th 2008, 05:58 AM
  4. Replies: 1
    Last Post: August 2nd 2006, 04:51 PM
  5. Didn't know to do this on test
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 6th 2006, 07:51 AM

Search Tags


/mathhelpforum @mathhelpforum