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 of a function

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data
    See the first photo.Q45

    Can I use the method similar to 43?
    I don't understand what does "condering (x,0,0)or (0,y,0) mean.


    2. Relevant equations



    3. The attempt at a solution
    see photo 2.
    I only know how to do 43
    I have no idea for 45
     

    Attached Files:

  2. jcsd
  3. Oct 1, 2011 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    For a limit to exist, you can't get different answers depending on the direction you approach a point. When the problem says to consider points of the form (x,0,0), it's asking you to calculate the limit as you approach the origin along the x-axis.
     
  4. Oct 1, 2011 #3

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    BTW: What do you conclude about #43?
     
  5. Oct 1, 2011 #4
    For 45 try this

    [tex]f(x,y,z) = \frac{x + y + z}{x^2 + y^2 + z^2}[/tex]

    What happens when you cut with x = x and with y,z =0? What happens to f?
     
    Last edited: Oct 1, 2011
  6. Oct 2, 2011 #5
    It becomes [itex]\frac{1}{x}[/itex]
    Then, as x[itex]\rightarrow[/itex]0, [itex]f(x,y,z)[/itex] doesn't exist?
     
  7. Oct 2, 2011 #6
    It seems if we have denominator, then the function is not continuous at origin.
     
  8. Oct 2, 2011 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    ??? The problem does not ask about continuity, only about the limit at the point. It is possible for the denominator of a fraction to go to 0 and the limit still exist at that point.

    In problem 43,
    [tex]\lim_{(x,y)\to (0,0)}\frac{x^2- y^2}{x^2+ y^2}[/tex]
    if we take y= mx, as suggested, we get
    [tex]\frac{x^2- m^2x^2}{x^2+m^2x^2}= \frac{x^2(1- m^2)}{x^2(1+ m^2)}= \frac{1- m^2}{1+ m^2}[/tex]
    as long as x is not 0.

    If the limit exists, then the limit of that should be the same for all m. Is it?
     
  9. Oct 2, 2011 #8
    It seems we use the similar method to prove continunity. If not, how do I prove that problem 43 continuous at (0,0)?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Limit of a function
  1. Limit of Functions (Replies: 1)

  2. Limit of a function (Replies: 7)

  3. Limit of a function (Replies: 7)

  4. Limits and functions (Replies: 16)

  5. Limit of function (Replies: 6)

Loading...