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Limits (another error in the question?)

  • Thread starter Yann
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  • #1
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Homework Statement



a) Prove that;

[tex]\lim_{(x,y)\rightarrow(0,0)} \frac{x^2y}{x^2+y^2} = 0[/tex]

b) Prove that if [tex]\lim_{(x,y)\rightarrow(0,0)} f(x,y) = L_1[/tex] and [tex]\lim_{(x,y)\rightarrow(0,0)} f(x,y) = L_2[/tex], then [tex]L_1=L_2[/tex]

c) Using the statement proven in 5b, prove that

[tex]\lim_{(x,y)\rightarrow(0,0)} \frac{xy}{x^2+y^2}[/tex]

Does NOT exist.

2. The attempt at a solution

a)

[tex]f(0,y) = \frac{0}{y^2} = 0[/tex]

[tex]f(x,0) = \frac{0}{x^2} = 0[/tex]

From those 2 directions, the limit is the same, so;

[tex]\lim_{(x,y)\rightarrow(0,0)} \frac{x^2y}{x^2+y^2} = 0[/tex]

b)

I have no idea how to do that, it seems to evident !

c)

Does NOT exist ? But...

[tex]f(0,y) = \frac{0}{y^2} = 0[/tex]

[tex]f(x,0) = \frac{0}{x^2} = 0[/tex]

It's exactly the same thing as in a), the limit DOES exist and it is;

[tex]\lim_{(x,y)\rightarrow(0,0)} \frac{xy}{x^2+y^2} = 0[/tex]
 

Answers and Replies

  • #2
Gib Z
Homework Helper
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Ahh for b), seeing as they are the same limit, they must evaluate to the same thing. L1=L2
 
  • #3
48
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ops, I got it wrong. The limit in c doesn't exist because at (x,mx);

[tex]\lim_{(x,y)\rightarrow(0,0)} \frac{mx^2}{x^2(1+m)} = \frac{m}{m+1}[/tex]

However, this leads me to 2 questions;

I just proved the limit in c doesn't exist, why on earth would I need, or even, how could I use the statement in b ?

Also, is my "proof" that the limit in a equal 0 sufficient ? Even at (x,mx) the limit is still 0, but still, is this a proof that the limit is 0 ?
 
Last edited:
  • #4
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The logic in problem c) is that since you have two different limits you contradicted b) so that the limit does not exist.

b) is just saying that if the (some) limit exists, then it is unique. Part b) can be resolved by general limit arguments, not talking just about function of two variables x,y.
 
  • #5
150
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You answer to a) does not account for (x,y) approaching (0,0) in any arbitrary path. What happens as the distance r from (0,0) goes to zero? (Hint: remember polar coords?)
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
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932
I think (b) is badly stated. I would interpret what they say as "the limit is unique". That's easy to prove. Suppose there existed two different limits, L1, L2. Take [itex]\epsilon= |L_1- L_2|/2[/itex]. Show that no [itex]\delta[/itex] makes both [itex]|f(x)- L_1|< \epsilon[/itex] and [itex]|f(x)- L_2|< \epsilon[/itex] for (x,y) such that [itex]\sqrt{(x^2+ y^2}< \delta[/itex].

However, that's clearly not what's meant. What they mean is that if the limit itself exists, then the limits along any path must be the same. Unfortunately, the way it is stated implies that the "limit" means the limit "along some path" which is not true.


As gammamcc said, showing that the limit along two different paths gives the same thing does not prove that the limit exists. you must show that f(x,y) is close to some L as long as (x,y) is close to (0,0). Again, as gammamcc said, consider changing to polar coordinates. That way r itself measures the distance from (0,0). If the limit, as r goes to 0, does not depend upon [itex]\theta[/itex], then the limit exists.
 
  • #7
48
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Great, thank you both, I have my answer.
 

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