# 0/0 DNE or undefined?

by negation
Tags: 0 or 0, undefined
 P: 819 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?
 Sci Advisor P: 839 Is that ##\frac{\sin(x)}{x} + y## or ##\frac{\sin(x)}{x+y}##?
Mentor
P: 18,346
 Quote by negation 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.
That is correct.

 What I've been hearing too is that since it is 0/0 for the above function, the limit DNE.
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.

P: 819
0/0 DNE or undefined?

 Quote by micromass 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.
Can I then presume a case of "loose" definition has been employed?

" the limiting behaviour is path dependent so lim of the function g(x,y) as (x,y) ->0 does not exists.
P: 819
 Quote by pwsnafu Is that ##\frac{\sin(x)}{x} + y## or ##\frac{\sin(x)}{x+y}##?
The former.

Edit: sorry, latter!

The former has a limit by performing l'hopital rule.
Mentor
P: 15,202
 Quote by negation 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.
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?
P: 819
 Quote by D H 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?
It is differentiable everywhere except x=0.
 Mentor P: 15,202 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.
P: 819
 Quote by D H 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.
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.
 Mentor P: 15,202 Continuity and limits go hand in hand. A function f(x) is continuous at some point x=a ifThe 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.

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