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!

0/0 DNE or undefined?

  1. Apr 29, 2014 #1
    What can we deduce about the lim g(x,y) as (x,y) -> (0,0)?
    where g(x,y) = sin(x)/x+y

    in substituiting, we get 0/0 so it has an indeterminate form which requires further work to ascertain if it is truly DNE or if it has a limit.
    What I've been hearing too is that since it is 0/0 for the above function, the limit DNE. Which is which? Or are definitions being loosely used?
  2. jcsd
  3. Apr 29, 2014 #2


    User Avatar
    Science Advisor

    Is that ##\frac{\sin(x)}{x} + y## or ##\frac{\sin(x)}{x+y}##?
  4. Apr 29, 2014 #3
    That is correct.

    That is incorrect. Just because you get a "0/0"-situation doesn't mean the limit does not exist. It does mean that you need to do some more work to find out what the limit is and whether it actually does exist.
  5. Apr 29, 2014 #4
    Can I then presume a case of "loose" definition has been employed?

    From my notes, it reads
    " the limiting behaviour is path dependent so lim of the function g(x,y) as (x,y) ->0 does not exists.
  6. Apr 29, 2014 #5
    The former.

    Edit: sorry, latter!

    The former has a limit by performing l'hopital rule.
  7. Apr 29, 2014 #6

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    That is the correct definition. In this case, ##\frac{\sin x}{x+y}## takes on different values as (x,y)→0 depending on the path. For example, the limit is 1 along the line y=0, but it's 1/2 along the line y=x. The limit does not exist.

    This can happen even in one dimension. What's the derivative of |x| at x=0?
  8. Apr 29, 2014 #7
    It is differentiable everywhere except x=0.
  9. Apr 29, 2014 #8

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Precisely. The one-sided limits ##\lim_{h \to 0^+} \frac{|x+h| - |x|}{h}## and ##\lim_{h \to 0^-} \frac{|x+h| - |x|}{h}## exist at x=0 but differ from one another. Therefore the two-sided limit ##\lim_{h \to 0} \frac{|x+h| - |x|}{h}## doesn't exist at x=0.
  10. Apr 29, 2014 #9
    I might be wrong. But intuitively, this appears to relate to the idea of continuity. From what you've stated, I gather that if both limit from the left f(x-) = f(x+) = f(x), then the graph is continuous.
  11. Apr 29, 2014 #10

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Continuity and limits go hand in hand. A function f(x) is continuous at some point x=a if
    • The function is defined at x=a (i.e., f(a) exists),
    • The limit of f(x) as x→a exists, and
    • These two quantities are equal to one another.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook