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

Question of basic calculus

  1. Sep 23, 2009 #1
    can someone please convince me that lim x->0 sqrt(x) = 0
    Who of you say it doesn't exist???
     
  2. jcsd
  3. Sep 23, 2009 #2

    statdad

    User Avatar
    Homework Helper

    If you mean

    [tex]
    \lim_{x \to 0^+} \sqrt{x}
    [/tex]

    (limit as x approaches 0 through the positives), then the limit is zero. Notice that this is a one-sided limit.

    This

    [tex]
    \lim_{x \to 0^-} \sqrt x
    [/tex]

    means you are trying to approach 0 from the left - through negative numbers. I feel confident in saying this limit does not exist. Why might that be?

    Finally, remember that the ordinary limit exists as a real number if, and only if, the two one-sided limits exist as real numbers and are equal. The comment and observation I made earlier combine to say (fill in the blank yourself) about

    [tex]
    \lim_{x \to 0} \sqrt x
    [/tex]
     
  4. Sep 23, 2009 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    HI Caesar_Rahil! :smile:

    lim x -> 0+ sqrt(x) = 0, but lim x->0- sqrt(x) depends on you define sqrt(x) for negative x …

    how are you defining it?

    Do you define it as not existing, or do you define it as an imaginary number?
     
  5. Sep 27, 2009 #4
    I'm confining myself to real valued functions
    since left hand limit does not exit, that means sqrt(x) does not have a limit there
    so it does not exist
    is this interpretation correct?
     
  6. Sep 27, 2009 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Yes … if you define sqrt(x) for negative x as not existing, then automatically there cannot be a limit at x = 0.

    (On the other hand, if you define sqrt as a function whose domain is the non-negative numbers only, then the limit does exist. :wink:)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question of basic calculus
Loading...