f(x, y) = x^2/2y. Prove the limit does not exist by using ε − δ definition.

statstudent123

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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 confused in proving this question since my calculus textbook does not provide any example in proving the case.
The ϵ,δ\epsilon,\delta formulation is (at least in L2L_2 metric) : Aϵ>0δ>0:x2+y2<δ2f(x,y)A<ϵ\exists A\forall \epsilon>0\exists\delta>0 : x^2+y^2<\delta^2\rightarrow |f(x,y)-A| < \epsilon. To prove the opposite you need to show that for any AA it is possible to find ϵ>0\epsilon>0 so that for any positive δ\delta a pair of x,yx,y can be found such x2+y2<δ2x^2+y^2<\delta^2 but (f(x,y)Aϵ|(f(x,y)-A| \geq \epsilon.

Usually it is easier to prove an equivalent statement: instead of proving the negation for any AA one can show that there exists an ϵ\epsilon such that for any δ\delta a pair of points x1,y1x_1,y_1 and x2,y2x_2,y_2 can be found inside the δ\delta-neighborhood such that f(x1,y1)f(x2,y2)ϵ|f(x_1,y_1)-f(x_2,y_2)| \geq \epsilon. Can you see that this is an equivalent statement?

P.S. An equivalent formulation of ϵ,δ\epsilon,\delta definition (in L1L_1 metric) would use x+y<δ|x|+|y|<\delta instead of x2+y2<δ2x^2+y^2<\delta^2. You should use the definition used in your class.
 
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.
 
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.
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\delta-neighborhood. But I don't see any contradiction yet. The fact that f(1,3)=1/6f(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\delta-neighborhood.
On the other hand, if you can prove that in any, however small, δ\delta-neighborhood one can find (x,y)(x,y) such that f(x,y)>2|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 do realize that f(x,y) = x^2/2y means f(x,y)=x2y2?\displaystyle f(x,y) = \dfrac{x^2y}{2}?

By any chance do you mean f(x,y)=x22y?\displaystyle f(x,y) = \dfrac{x^2}{2y}? If yes, then you should write x^2/(2y). Those parenthesis are important.

 
You do realize that f(x,y) = x^2/2y means f(x,y)=x2y2?\displaystyle f(x,y) = \dfrac{x^2y}{2}?

By any chance do you mean f(x,y)=x22y?\displaystyle f(x,y) = \dfrac{x^2}{2y}? If yes, then you should write x^2/(2y). Those parenthesis are important.

Yes sir.
 
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\delta-neighborhood. But I don't see any contradiction yet. The fact that f(1,3)=1/6f(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\delta-neighborhood.
On the other hand, if you can prove that in any, however small, δ\delta-neighborhood one can find (x,y)(x,y) such that f(x,y)>2|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.
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?
 
You have to prove that the relevant (x,y)(x,y) can be found for any δ\delta.
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.
 
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