Results 1 to 7 of 7

Math Help - limit - more about division

  1. #1
    Newbie
    Joined
    Jan 2006
    Posts
    16

    limit - more about division

    lim ((1/sqrtx)-1)/(x-1)
    x>1

    (1-sqrtx/sqrtx)/(x-1)

    why does this end up looking like this (1-sqrtx)/(sqrtx(x-1))

    i am having a brain fart need this for my exam
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    \lim_{x\rightarrow{1}}\frac{\frac{1}{\sqrt{x}}-1}{x-1}

    As you said, this is equivalent to \lim_{x\rightarrow{1}}\frac{1-\sqrt{x}}{x^\frac{3}{2}-\sqrt{x}}

    Now would be a good time to use L'Hopital's rule, if you have learned this. Have you?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by tnkfub
    lim ((1/sqrtx)-1)/(x-1)
    x>1

    (1-sqrtx/sqrtx)/(x-1)
    why does this end up looking like this (1-sqrtx)/(sqrtx(x-1))
    i am having a brain fart need this for my exam
    Hello,

    here is a way to solve this problem without using the l'Hospital rule:

    Substitute the x by (1+1/n) and calculate the limit for n approaching infinity:

    \lim_{\csub{n \rightarrow \infty}} {\left( \frac{ \frac{1}{ \sqrt{1+\frac{1}{n}}}-1}{1+\frac{1}{n} - 1}  \right)}

    After a few steps of very unpleasant transformation you'll get:

    \lim_{\csub{n \rightarrow \infty}} {\left( \frac{n \cdot \left( n \cdot  \sqrt{\frac{n+1}{n}}-n-1  \right)}{n+ 1}  \right)}=- \frac{1}{2}


    Greetings

    EB
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by tnkfub
    lim ((1/sqrtx)-1)/(x-1)
    x>1

    (1-sqrtx/sqrtx)/(x-1)

    why does this end up looking like this (1-sqrtx)/(sqrtx(x-1))

    i am having a brain fart need this for my exam
    You have,
    \frac{\frac{1}{\sqrt{x}}-1}{x-1}
    You can express as,
    \frac{\frac{\sqrt{x}}{x}-1}{x-1}=\frac{\sqrt{x}-x}{x(x-1)}
    Thus,
    \frac{\sqrt{x}-x}{x(x-1)}\cdot \frac{\sqrt{x}+x}{\sqrt{x}+x}
    Thus,
    \frac{x-x^2}{x(x-1)(\sqrt{x}+x)}
    Thus,
    \frac{-x(x-1)}{x(x-1)(\sqrt{x}+x)}
    Thus,
    -\frac{1}{\sqrt{x}+x}
    As, x\to 1 you have,
    -\frac{1}{\sqrt{1}+1}=-\frac{1}{2}
    Last edited by ThePerfectHacker; April 22nd 2006 at 05:22 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by earboth
    Hello,

    here is a way to solve this problem without using the l'Hospital rule:

    Substitute the x by (1+1/n) and calculate the limit for n approaching infinity:

    \lim_{\csub{n \rightarrow \infty}} {\left( \frac{ \frac{1}{ \sqrt{1+\frac{1}{n}}}-1}{1+\frac{1}{n} - 1}  \right)}

    After a few steps of very unpleasant transformation you'll get:

    \lim_{\csub{n \rightarrow \infty}} {\left( \frac{n \cdot \left( n \cdot  \sqrt{\frac{n+1}{n}}-n-1  \right)}{n+ 1}  \right)}=- \frac{1}{2}


    Greetings

    EB
    Can you prove that such a substitution is valid?
    Not that I am arguing I am simply interested.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by ThePerfectHacker
    Can you prove that such a substitution is valid?
    Not that I am arguing I am simply interested.
    Hello,

    honestly speaking: No.

    1. When you calculate the derivative you calculate for instance the limit dx = \lim_{\csub{x \rightarrow x_0}}{(x_0-x)}= \lim_{\csub{n \rightarrow \infty}}{(x_0-(x_0+\frac{1}{n}))

    2. I made an analoguous conclusion to use this method with this problem. I noticed that it worked fine. But of course that isn't a proof.

    Greetings

    EB
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by earboth
    Hello,

    honestly speaking: No.

    1. When you calculate the derivative you calculate for instance the limit dx = \lim_{\csub{x \rightarrow x_0}}{(x_0-x)}= \lim_{\csub{n \rightarrow \infty}}{(x_0-(x_0+\frac{1}{n}))

    2. I made an analoguous conclusion to use this method with this problem. I noticed that it worked fine. But of course that isn't a proof.

    Greetings

    EB
    Not arguing with what you are saying, and I would use it if I was taking an exam without knowing its proof. I was just always curious about why we can do that. Seen it used many times.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Division
    Posted in the Algebra Forum
    Replies: 3
    Last Post: July 7th 2011, 09:45 PM
  2. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 3rd 2009, 05:05 PM
  3. Polynomial division vs synthetic division
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 9th 2009, 05:49 AM
  4. Division: The big 7 way
    Posted in the Algebra Forum
    Replies: 10
    Last Post: December 15th 2008, 01:29 AM
  5. division
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: November 16th 2008, 09:57 PM

Search Tags


/mathhelpforum @mathhelpforum