statstudent123
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- Jan 18, 2022
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I am confused in proving this question since my calculus textbook does not provide any example in proving the case.
I am guessing x,y→0.What limit? (x, y) -> ?
The ϵ,δ formulation is (at least in L2 metric) : ∃A∀ϵ>0∃δ>0:x2+y2<δ2→∣f(x,y)−A∣<ϵ. To prove the opposite you need to show that for any A it is possible to find ϵ>0 so that for any positive δ a pair of x,y can be found such x2+y2<δ2 but ∣(f(x,y)−A∣≥ϵ.I am confused in proving this question since my calculus textbook does not provide any example in proving the case.
Consider arbitrarily small δ-neighborhood and its intersection with the curve y=x2.I think the limit approaches 0, so it does exist.
lim(x, y) approach to (0, 0)What limit? (x, y) -> ?
This does not look like a proof to me. You've proved that if a limit L exists than -1 < L < 1, and that |(x^2/2y)|< 2 for (x,y) in the δ-neighborhood. But I don't see any contradiction yet. The fact that f(1,3)=1/6 does not prove anything to me. Moreover, you don't even know whether the (1,3) point is inside the δ-neighborhood.I am trying to solve the problem by using contradiction, but I am not sure whether it is true.
Let lim (x, y) approach (0, 0) (x^2/2y) =L and suppose the limit exist. Then we let ϵ=1 and let δ>0 such that if (x, y) is the domain and 0<sqrt(x^2+y^2)<δ, then lim (x, y) approach (0, 0) |(x^2/2y)-L|<ϵ. Then, lim (x, y) approach (0, 0) |(x^2/2y)-L|<1.
Note that -1<|(x^2/2y)-L|<1, it follows that L-1<(x^2/2y)<L+1. Since |(x^2/2y)| have both positive and negative values, then L-1<0 and l+1>0. It follows that -1<L<1. Then, |(x^2/2y)|<L+1<1+1=2.
Now, we let (x, y)=(1, 3). We get |1^2/2(3)|=1/6 is not equal to 2. Therefore, the limit does not exist.
Sorry but |1^2/2(3)|= 3/2Now, we let (x, y)=(1, 3). We get |1^2/2(3)|=1/6 is not equal to 2. Therefore, the limit does not exist.
Yes sir.You do realize that f(x,y) = x^2/2y means f(x,y)=2x2y?
By any chance do you mean f(x,y)=2yx2? If yes, then you should write x^2/(2y). Those parenthesis are important.
How if I let (x, y)= (3, 2). Then 0<sqrt(3^2+2^2)=sqrt(13)<δ, but 3^2/2(2)=2.25>2, which gives a contradiction?This does not look like a proof to me. You've proved that if a limit L exists than -1 < L < 1, and that |(x^2/2y)|< 2 for (x,y) in the δ-neighborhood. But I don't see any contradiction yet. The fact that f(1,3)=1/6 does not prove anything to me. Moreover, you don't even know whether the (1,3) point is inside the δ-neighborhood.
On the other hand, if you can prove that in any, however small, δ-neighborhood one can find (x,y) such that ∣f(x,y)∣>2 it will contradict the statement you've proved earlier, thus completing the proof by contradiction.
P.S. There are also some typo in your text, like "|(x^2/2y)| have both positive and negative values" -- I am sure you did not mean the absolute value, which can never be negative.
You have to prove that the relevant (x,y) can be found for any δ.How if I let (x, y)= (3, 2). Then 0<sqrt(3^2+2^2)=sqrt(13)<δ, but 3^2/2(2)=2.25>2, which gives a contradiction?
Now I let x=δ/2 and y =δ^2/20. We have 0<=sqrt(x^2+y^2)=sqrt((δ/2)^2+(δ^2/20)^2)=δ/2 sqrt(1+δ^2/100)<δ,You have to prove that the relevant (x,y) can be found for any δ.
I like this one much more!Now I let x=δ/2 and y =δ^2/20. We have 0<=sqrt(x^2+y^2)=sqrt((δ/2)^2+(δ^2/20)^2)=δ/2 sqrt(1+δ^2/100)<δ,
but |x^2/(2y)|=|(δ/2)^2/2(δ^2/20)|=2.5>2, which gives a contradiction.
Oh yeah. Thanks for your help?I like this one much more!