Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Lim as X approaches 2 (rationalizing wrong)

  1. Sep 4, 2007 #1
    1. The problem statement, all variables and given/known data

    The limit as x approaches 2 for (X-2)/(sqrt7+x)-(x+1)

    2. Relevant equations



    3. The attempt at a solution
    I know i have to rationalize the denominator but it seems like i'm doing something very wrong with my distrubution...please help!!!
     
  2. jcsd
  3. Sep 4, 2007 #2

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    are X and x supposed to be distinct?

    When writing things like this out, it's worth being a bit clearer, since what you've written could be:
    [tex]\frac{X-2}{\sqrt{7}+x}-(x+1)[/tex]
    or
    [tex]\frac{X-2}{(\sqrt{7}+x)-(x+1)}[/tex]

    Regardless, I don't see why you would need to rationalize the denominator.
     
  4. Sep 4, 2007 #3
    lim as x approaches 2 for [tex]X-2/\sqrt{7+X}-(x+1)[/tex]

    this was the problem..I solved it to be -6/5......I just have a question...i have another similar problem, which also becomes in the indeterminate form.

    lim as t approaches 3 for [tex]1-t+\sqrt{1+t}/t-3[/tex]

    where do i find more information on how to distribute here? I know i have to rationalize I am just lost in how distribution works with a problem with no parentheses and one with parentheses?

    I have 4 different books here, 2 algebra, 2 calculus....and not sure what or where to review this
     
  5. Sep 4, 2007 #4

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    In latex the construct for fractions is:
    \frac{$numerator}{$denominator}
    (You can click on the graphical version to see the code:
    [tex]\frac{1}{4}[/tex]
    It will make things a bit more legible.

    You seem to be using [itex]X[/itex]and [itex]x[/itex] as if they were the same - they're not.

    To rationalize:
    [tex]\frac{x-2}{\sqrt{7+x}-(x+1)}[/tex]
    Multiply by:
    [tex]\frac{\sqrt{7+x}+(x+1)}{\sqrt{7+x}+(x+1)}[/tex]

    Generally, if you have:
    [tex]\sqrt{a} + b[/tex]
    you'll want to multiply by
    [tex]\sqrt{a} - b [/tex]
    since this creates a difference of two squares:
    [tex](\sqrt{a} + b)\times(\sqrt{a} - b )=a-b^2[/tex]
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook