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: Simple Limit Problem

  1. Jun 21, 2006 #1

    es

    User Avatar

    I tutor high school students in Calc and the other day I came across this problem.

    Limit of (1-Sqrt(x-2))/(x-3) as x->3

    I tried coaching the student on how to simplify the expression and in the end I just showed him this substitution.

    Let u=Sqrt(x-2)

    Then

    (1-Sqrt(x-2))/(x-3) = (1-u)/(u^2-1)

    And the Limit becomes

    Limit of (1-u)/(u^2-1)=-1/(u+1) as u->1 which is -1/2

    He looked at me like I had just done some black magic. I explained substitution to him and why it worked, showed him a couple of other simple examples, and confirmed the answer numerically (like they do in basic calc books when the limit concept is first presented). I still don't think he is 100% convinced because they had not covered this in his class yet which leads to my question.

    Can the original problem be solved without substitution?
     
  2. jcsd
  3. Jun 21, 2006 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Well, you might do it this way:
    [tex]\frac{(1-\sqrt{x-2})}{x-3}=\frac{(1-\sqrt{x-2})}{x-3}*1=\frac{(1-\sqrt{x-2})}{x-3}*\frac{(1+\sqrt{x-2})}{(1+\sqrt{x-2})}=\frac{3-x}{(1+\sqrt{x-2})*(x-3)}=-\frac{1}{(1+\sqrt{x-2})}[/tex]
    And so on..
     
  4. Jun 21, 2006 #3
    Multiplying by the conjugate would do the trick, but personally, I feel the easiest way is L'Hospital's rule .
     
  5. Jun 22, 2006 #4

    es

    User Avatar

    Yipe. For some reason I though 1-x would be the numerator. Boy do I feel sheepish. :) Thanks for the help!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook