1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit in the complex plane

  1. Feb 13, 2017 #1
    1. The problem statement, all variables and given/known data
    lim as z--> i , [itex] \frac{z^2-1}{z^2+1} [/itex]
    3. The attempt at a solution
    When we plug in i we get -2/0, so we get division by 0, Does this mean the limit is
    infinity, I also tried approaching from z=x+i where x went to 0, you get the same answer,
    I also approached from z=yi where y approaches 1, and I got the same answer.
     
    Last edited by a moderator: Feb 13, 2017
  2. jcsd
  3. Feb 13, 2017 #2

    Mark44

    Staff: Mentor

    What happens with ##\lim_{x \to 0} \frac 1 x##? Is the limit here ##\infty## or does the limit simply not exist at all? IOW, are the two one-sided limits equal?
     
  4. Feb 13, 2017 #3
    good point if we appraoch x from the left it goes to minus infinity , and if we approach from the right it is positive infinity, so the limit does not exist.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Limit in the complex plane
  1. Complex Limits (Replies: 1)

  2. Complex limit (Replies: 3)

  3. Complex Limit (Replies: 3)

  4. Complex Limits (Replies: 3)

  5. Complex Limit (Replies: 4)

Loading...