Is there a function that satisfies these two limit conditions?

[kleinwolf]

I'm searching for a function f(x,y), such that

1) $$\lim_{t\rightarrow 0} f(at,bt)=E \quad\forall(a,b)\ne(0,0)$$
2) $$\exists a,b|\lim_{t\rightarrow 0}f(at,bt^2)=F\ne E$$

 

[Hurkyl]

Sounds like homework. Anyways, what thoughts have you had so far?

 

[kleinwolf]

I think it's impossible, but I wanted to have other people opinion. I will tell you why :

take f(x,y) and just use polar coordinates : x=r*cos(theta), y=r*sint(theta). Just put : t=r, cos(theta)=a, sin(theta)=b. Then from 1), the limit as (x,y)->(0,0) exists and is E. Hence, it should be independent on how you tend towards (0,0).

Since b) is a special case of the limit (x,y)->(0,0), then it should be equal to E.

But I'm not sure if this reasoning is correct.

 

[Hurkyl]

You have it backwards -- if you show the limit doesn't depend on how you approach 0, then you can conclude the limit exists.

All you've shown is that if the limit exists, it is E.

 

[BicycleTree]

I don't think I know much about this type of math, but the restriction (a, b) != (0,0) only applies to 1). In 2), (a, b) can equal (0, 0), and you can use that.

 

[Hurkyl]

You don't have to "cheat" like that, though. In fact, I think you can change part 2 to say "For all a and b", but the problem then becomes more difficult.

Anyways, you try converting part 2 to polar coordinates too?

 

[kleinwolf]

Yes Bicycletree is right : (a,b) should be different than (0,0) in b...

In fact the question can be simply stated in words : if you show that approaching (0,0) on every straight line leads to the same limit, then can we deduce this is the limit approaching whatever way you want (on curves)...I think I could just say : a curve, when becoming near to 0 can be approximated by a straight line ?

 

[snoble]

I don't want to answer your question exactly but I want you to consider the following polar function $$f(r, \theta) = r/\theta$$ where $$0 < \theta \le 2\pi$$. Certainly if you fix a theta the limit as r approaches 0 is 0. But if you take the path $$g(t) = (1/t,1/t)$$. Then $$f\circ g(t) =1$$ for t>1. But the path is spiraling towards the origin since r is approaching 0. This is a cooked example that won't satisfy your question but first try creating a cooked example that will satisfy your question (something piecewise... this is easy) then if you are feeling adventurous you can come up with some a little less cooked looking (harder).

good luck
Steven

 

[matt grime]

Yes Bicycletree is right : (a,b) should be different than (0,0) in b...

In fact the question can be simply stated in words : if you show that approaching (0,0) on every straight line leads to the same limit, then can we deduce this is the limit approaching whatever way you want (on curves)...I think I could just say : a curve, when becoming near to 0 can be approximated by a straight line ?

But once more you're assuming that f must be continuous (at 0,0) to show that the result is false, surely that should tell you where to look for a counter example?

The common way to do this kind of question is to make f(x,y) a function of the form g(x,y)/h(x,y) where you can in the first case pull out factor of t so that we're left with f(ta,tb)=tf(a,b) but if we put in (at,bt^2) we cancel all the factors of t so that f(at,bt^2)=f(a,b) and the limit is independent of t (but such that f is not constant).

Examples that don't quite work here:

f(x,y) = x^2/y

f(ta,tb)=ta^2/b which tends to zero as t tends to zero

f(ta,t^2b)= a^2/b

of course this doesn't quite work since b could be zero. I leave it to you to sidestep this problem.