Results 1 to 12 of 12

Thread: 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?

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

  2. #2
    Super Member
    Joined
    Jul 2012
    From
    INDIA
    Posts
    863
    Thanks
    220

    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
    2,005
    Thanks
    747

    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
    2,005
    Thanks
    747

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

    Yes, and as $\displaystyle x\to\infty$ 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 $\displaystyle x\to\infty$ the limit is 1/2.
    and what about the limit as $\displaystyle x\rightarrow-\infty$ , 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
    2,005
    Thanks
    747

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

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

    $\displaystyle \lim_{x\to\infty}\left(\sqrt{x^2-x}+x \right)$
    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
    2,005
    Thanks
    747

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

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

  10. #10
    Senior Member zzephod's Avatar
    Joined
    Apr 2012
    From
    Erewhon
    Posts
    261
    Thanks
    153

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

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

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

    rather than:

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

    If you want to try going down that route.

    .
    Last edited by zzephod; Apr 26th 2013 at 02:54 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
    21,736
    Thanks
    2811
    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 $\displaystyle \lim_{x\rightarrow{\color{red}-}\infty}(\sqrt{x^{2}+x}-x) = ? $

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

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

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Oct 23rd 2012, 11:00 AM
  2. Replies: 2
    Last Post: Nov 8th 2009, 02:18 PM
  3. Several questions I didn't get
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 6th 2008, 04:58 AM
  4. Replies: 1
    Last Post: Aug 2nd 2006, 03:51 PM
  5. Didn't know to do this on test
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Jul 6th 2006, 06:51 AM

Search Tags


/mathhelpforum @mathhelpforum