Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - limit with restricted ways of doing it

  1. #1
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1

    limit with restricted ways of doing it

    Hey Gang:

    Here is a limit I saw somewhere recently but can not remember where. I think on another site.

    Anyway, the restrictions are no expansions on e, no binomial expansion on the radical term and no L'Hopital. Go figure.

    \lim_{x\to 0}\frac{e^{x}-x-\sqrt{1+x^{2}}}{x^{3}}

    I just thought it'd be cool to see some clever ways folks come up with.
    Last edited by galactus; September 26th 2008 at 04:42 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Someone posted it here a few days ago but got no answers:

    http://www.mathhelpforum.com/math-he...expansion.html

    The answer still eludes me.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    The solution is 1/6
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by galactus View Post
    The solution is 1/6
    Maple tells me that \lim_{x\to 0}\frac{e^{x}-x-\sqrt{1-x^{2}}}{x^{3}} is undefined...

    --Chris
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by Chris L T521 View Post
    Maple tells me that \lim_{x\to 0}\frac{e^{x}-x-\sqrt{1-x^{2}}}{x^{3}} is undefined...

    --Chris
    It should be \sqrt{1+x^2}

    EDIT: attached a graph
    Attached Thumbnails Attached Thumbnails limit with restricted ways of doing it-picture-1.png  
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Yes, I just noticed that and changed my typo. Sorry about that.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    It's been posted too in Mathlinks, but answers do include Taylor expansions content. (At least, we avoided L'H˘pital's Rule.)

    Anyway, it's kinda hard, I can't see a solution yet.

    May be someone else's gonna give a nice solution.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    This same question was also posted here.

    --Chris
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Could the squeeze theorem be applied to this beast??? Or would it turn out to be very tedious?? If we do find a way to evaluate this with the given restrictions, it will be a tedious process any...unless someone is very clever...

    --Chris
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    By taking derivatives we can prove on (-1,1) we have:
    1+x+\frac{x^2}{2}+\frac{x^3}{6} \leq e^x \leq 1 + x + \frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12}

    (These inequalities are inspired by Talyor's theorem but they make no mention of the theorem).

    Uses these inequalities we can squeeze the function. The limit is completed by rationalizing the numerator. Here is an illustration with the lower bound.

    \frac{\left( 1 + x+\frac{x^2}{2}+\frac{x^3}{6}\right) - x - \sqrt{x^2+1}}{x^3} \leq \frac{e^x - x - \sqrt{x^2+1}}{x^3}

    Now this is,
    \frac{\left( 1 + \frac{x^2}{2} + \frac{x^3}{6} \right) - \sqrt{x^2+1}}{x^3} \cdot \frac{\left( 1 + \frac{x^2}{2}+\frac{x^3}{6} \right) + \sqrt{x^2+1}}{\left( 1 + \frac{x^2}{2}+\frac{x^3}{6} \right) + \sqrt{x^2+1}} = \frac{\left( 1 + \frac{x^2}{2} + \frac{x^3}{6} \right)^2 - x^2 - 1 }{x^3 \left( 1 + \frac{x^2}{2} + \frac{x^3}{6} + \sqrt{x^2+1} \right)}

    When we expand the trinomial we get 1+\frac{x^2}{2}+\frac{x^2}{2} + \frac{x^3}{6} + \text{ higher degree terms}

    After canceling and dividing by x^3 and ingoring higher degree terms we get \frac{1}{6} as a lower bound.

    Now do the same thing with the second polynomial.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    317
    The limit does not change if x is replaced by -x

    i.e. L=\lim_{x\to 0}\frac{e^x-x-\sqrt{1+x^2}}{x^3}

    L=\lim_{x\to 0}\frac{-e^{-x}-x+\sqrt{1+x^2}}{x^3}

     <br />
L=\lim_{x\to 0}\frac{e^x-e^{-x}-2x}{2x^3}<br />

    Putting x = 3t

    L = \lim_{t\to 0}\frac {e^{3t} - e^{ - 3t} - 6 t }{54 t^3}

    54L = \lim_{t\to 0}\frac {(e^{t} - 1)^3 + (3e^{2t} - 3e^{t} + 1) - (e^{ - t} - 1)^3 - 3e^{ - 2t} + 3e^{ - t} - 1 - 6 t }{ t^3}

     <br />
54L = \lim_{t\to 0}(\frac {e^t - 1}{t})^3 + (\frac {e^{ - t} - 1}{ - t})^3 + \frac {3e^{2t} - 3e^{ - 2t} - 3e^{t} + 3e^{ - t} - 6t}{t^3}<br />

     <br />
54L = \lim_{t\to 0}(\frac {e^t - 1}{t})^3 + (\frac {e^{ - t} - 1}{ - t})^3 + 48\frac {e^{2t} - e^{ - 2t} - 4t}{2(2t)^3} - 6\frac {e^t - e^{ - t} - 2t}{2(t)^3}<br />

    54L = 1 + 1 + 48L - 6L = 2 + 42L

     <br />
12L = 2<br />

     <br />
L = \frac {1}{6}<br />
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by pankaj View Post
    The limit does not change if x is replaced by -x
    i.e. L=\lim_{x\to 0}\frac{e^x-x-\sqrt{1+x^2}}{x^3}
    There is still a problem. You are assuming the limit exists.
    You need to show the limit actually exists. (Perhaps you can use fact that this is a decreasing function somehow).
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Who missed this one!

    \lim_{x\to{0}}\frac{e^x-x-\sqrt{1+x^2}}{x^3}

    Let \varphi=x^3

    Then this becomes

    \begin{aligned}\lim_{\varphi\to{0}}\frac{e^{\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}}{\varphi}&=\lim_{\varphi\to{0}}\frac{e^  {\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}-0}{\varphi-0}\\<br />
&=\left(e^{\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}\right)'\bigg|_{x=0}\\<br />
&=\frac{1}{6}\end{aligned}

    You did just say no L'hopitals or Series.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Mathstud28 View Post
    Who missed this one!

    \lim_{x\to{0}}\frac{e^x-x-\sqrt{1+x^2}}{x^3}

    Let \varphi=x^3

    Then this becomes

    \begin{aligned}\lim_{\varphi\to{0}}\frac{e^{\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}}{\varphi}&=\lim_{\varphi\to{0}}\frac{e^  {\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}-0}{\varphi-0}\\<br />
&=\left(e^{\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}\right)'\bigg|_{x=0}\\<br />
&=\frac{1}{6}\end{aligned}

    You did just say no L'hopitals or Series.


    Hahaha...clever substitution!! How long did it take you to figure it out?

    --Chris
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Chris L T521 View Post


    Hahaha...clever substitution!! How long did it take you to figure it out?

    --Chris
    I dont know. Like a minute and a half?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Restricted outcomes for die thrown six times.
    Posted in the Statistics Forum
    Replies: 4
    Last Post: July 17th 2011, 04:40 PM
  2. [SOLVED] Equivalent Integral for restricted values
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 13th 2011, 10:33 AM
  3. Inverse With Restricted Domain
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 19th 2009, 10:28 AM
  4. knight's restricted walk
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: September 12th 2009, 05:34 AM
  5. restricted command found (def)
    Posted in the LaTeX Help Forum
    Replies: 0
    Last Post: October 9th 2008, 09:27 AM

Search Tags


/mathhelpforum @mathhelpforum