Conditions Needed for Interchange of Double Limits

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What conditions must f satisfy if

[tex]\lim_{x \to a} \lim_{y \to b} f(x,y)=\lim_{y \to b} \lim_{x \to a} f(x,y)[/tex]

where [tex]\lim_{x \to a} f(x,y)[/tex] and [tex]\lim_{y \to b} f(x,y)[/tex] exists and are finite?
 
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Without condition you're statement is not true, consider
[tex]\frac{y^2}{x^2+y^2}.[/tex]
 
Jarle said:
What conditions must f satisfy if

[tex]\lim_{x \to a} \lim_{y \to b} f(x,y)=\lim_{y \to b} \lim_{x \to a} f(x,y)[/tex]

where [tex]\lim_{x \to a} f(x,y)[/tex] and [tex]\lim_{y \to b} f(x,y)[/tex] exists and are finite?

Good question. I'm fairly sure that continuity in both x and y would be a sufficient condition.
 
f should be continuous at (a,b) ... that is, continuous as a function of two variables. Continuous separately in each of x and y is not enough.
 
g_edgar said:
f should be continuous at (a,b) ... that is, continuous as a function of two variables. Continuous separately in each of x and y is not enough.

Arh yes, thanks for the clarification g_edgar. :)

BTW, what is the simplist definition of continuity in this case. I was thinking of something like :

[itex]\exists \, \, \epsilon > 0 \, : \, |f(x+dx,y+dy) \, - \, f(x,y) | \leq \, k \, ||(dx,dy)|| [/tex] whenever [itex]||(dx,dy)|| \leq \epsilon[/itex].<br /> <br /> Is that basically correct?[/itex]
 
If [tex]f[/tex] is continuous the we certainly have [tex]\lim_{x\rightarrow a} \lim_{y\rightarrow b} f(x,y) = \lim_{y\rightarrow b} \lim_{x\rightarrow a} f(x,y).[/tex]
But you do not need that much. Consider
[tex]g(x,y) = \frac{xy}{x^2+y^2}.[/tex]
The function is discontinuous at [tex](0,0)[/tex], since [tex]\lim_{t\rightarrow 0} g(t,t) = 1/2 \neq 0 = \lim_{t\rightarrow 0} g(t,0)[/tex].
But we have [tex]\lim_{x\rightarrow 0} \lim_{y\rightarrow 0} g(x,y) = \lim_{y\rightarrow 0} \lim_{x\rightarrow 0} g(x,y)=0,[/tex] and [tex]\lim_{x\rightarrow 0} g(x,y) = 0 = \lim_{y\rightarrow 0} g(x,y)[/tex] exist and are finite.
 
uart said:
Arh yes, thanks for the clarification g_edgar. :)

BTW, what is the simplist definition of continuity in this case. I was thinking of something like :

[itex]\exists \, \, \epsilon > 0 \, : \, |f(x+dx,y+dy) \, - \, f(x,y) | \leq \, k \, ||(dx,dy)|| [/tex] whenever [itex]||(dx,dy)|| \leq \epsilon[/itex].<br /> <br /> Is that basically correct?[/itex]
[itex] <br /> No that's not it. Try again![/itex]
 
Thanks for the replies.

Now, consider [tex]f(x,y)=x^y[/tex] on [tex](0,\infty)[/tex] in both variables. Is the function is continuous in both variables on the interval, and not only separately? We have existing limits as [tex]x[/tex] and [tex]y \to 0[/tex] independently. They are 0 and 1 respectively. However, the resulting limit depends on the order of the limit composition.

How do you account for this example? What conditions do f fail to satisfy? And what is the difference between continuity in two variables, and continuity in two variables separately?
 
Jarle said:
Thanks for the replies.

Now, consider [tex]f(x,y)=x^y[/tex] on [tex](0,\infty)[/tex] in both variables. Is the function is continuous in both variables on the interval, and not only separately?

Yes, [itex]f[/itex] is continuous on the whole product set [itex](0,\infty) \times (0,\infty)[/itex].

We have existing limits as [tex]x[/tex] and [tex]y \to 0[/tex] independently. They are 0 and 1 respectively. However, the resulting limit depends on the order of the limit composition.

How do you account for this example? What conditions do f fail to satisfy? And what is the difference between continuity in two variables, and continuity in two variables separately?

The condition that fails: [itex]f[/itex] is not continuous at the point [itex](0,0)[/itex]

Remember what I said back there? "f should be continuous at (a,b)" ?
 
A function f is continuous at a point v, if for every e>0 there is a d>0 such that
||f(v + u) - f(v)|| < e whenever ||u|| < d.