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0/0 DNE or undefined?

by negation
Tags: 0 or 0, undefined
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negation
#1
Apr29-14, 04:46 AM
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?
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pwsnafu
#2
Apr29-14, 05:34 AM
Sci Advisor
P: 834
Is that ##\frac{\sin(x)}{x} + y## or ##\frac{\sin(x)}{x+y}##?
micromass
#3
Apr29-14, 06:17 AM
Mentor
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Quote Quote by negation View Post
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.

negation
#4
Apr29-14, 06:43 AM
P: 819
0/0 DNE or undefined?

Quote Quote by micromass View Post
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?

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.
negation
#5
Apr29-14, 06:43 AM
P: 819
Quote Quote by pwsnafu View Post
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.
D H
#6
Apr29-14, 09:10 AM
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P: 15,170
Quote Quote by negation View Post
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?
negation
#7
Apr29-14, 09:33 AM
P: 819
Quote Quote by D H View Post
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.
D H
#8
Apr29-14, 09:40 AM
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P: 15,170
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.
negation
#9
Apr29-14, 09:50 AM
P: 819
Quote Quote by D H View Post
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.
D H
#10
Apr29-14, 10:29 AM
Mentor
P: 15,170
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.


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