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Homework Help: 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


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    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


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    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


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    ??? 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)?
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