Is there a function that satisfies these two limit conditions?

  • Thread starter kleinwolf
  • Start date
  • Tags
    Limit
In summary, the conversation discusses the search for a function f(x,y) that satisfies certain limits and examines the possibility of using polar coordinates to find a solution. There is a debate about the restrictions on the limits and the continuity of the function at (0,0). Examples are provided to illustrate the points being discussed.
  • #1
kleinwolf
295
0
I'm searching for a function f(x,y), such that

1) [tex] \lim_{t\rightarrow 0} f(at,bt)=E \quad\forall(a,b)\ne(0,0) [/tex]
2) [tex] \exists a,b|\lim_{t\rightarrow 0}f(at,bt^2)=F\ne E [/tex]
 
Physics news on Phys.org
  • #2
Sounds like homework. Anyways, what thoughts have you had so far?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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?
 
  • #7
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 ?
 
  • #8
I don't want to answer your question exactly but I want you to consider the following polar function [tex] f(r, \theta) = r/\theta [/tex] where [tex]0 < \theta \le 2\pi[/tex]. Certainly if you fix a theta the limit as r approaches 0 is 0. But if you take the path [tex]g(t) = (1/t,1/t) [/tex]. Then [tex]f\circ g(t) =1[/tex] 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
 
  • #9
kleinwolf said:
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.
 

FAQ: Is there a function that satisfies these two limit conditions?

1. What are the two limit conditions that the function must satisfy?

The two limit conditions are the left-hand limit and the right-hand limit. The left-hand limit is the value that the function approaches as the input approaches a specific value from the left side. The right-hand limit is the value that the function approaches as the input approaches a specific value from the right side.

2. How do I know if a function satisfies these two limit conditions?

A function satisfies the two limit conditions if the left-hand limit and the right-hand limit are equal to each other at the specific value. This means that the function approaches the same value from both sides at that specific value.

3. Can a function have different limit conditions at different points?

Yes, a function can have different limit conditions at different points. This means that the function may approach different values from both the left and right sides at different points.

4. What happens if a function does not satisfy these two limit conditions?

If a function does not satisfy the two limit conditions, it is not considered a continuous function. This means that there is a discontinuity at the specific value, and the function does not approach the same value from both sides.

5. Are there any restrictions on the type of function that can satisfy these two limit conditions?

No, there are no restrictions on the type of function that can satisfy these two limit conditions. As long as the function approaches the same value from both sides at the specific value, it satisfies the two limit conditions.

Similar threads

Replies
9
Views
2K
Replies
3
Views
2K
Replies
4
Views
2K
Replies
6
Views
1K
Replies
3
Views
947
Replies
1
Views
1K
Back
Top