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

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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.
 
statstudent123 said:
I am confused in proving this question since my calculus textbook does not provide any example in proving the case.
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The [imath]\epsilon,\delta[/imath] formulation is (at least in [imath]L_2[/imath] metric) : [imath]\exists A\forall \epsilon>0\exists\delta>0 : x^2+y^2<\delta^2\rightarrow |f(x,y)-A| < \epsilon[/imath]. To prove the opposite you need to show that for any [imath]A[/imath] it is possible to find [imath]\epsilon>0[/imath] so that for any positive [imath]\delta[/imath] a pair of [imath]x,y[/imath] can be found such [imath]x^2+y^2<\delta^2[/imath] but [imath]|(f(x,y)-A| \geq \epsilon[/imath].

Usually it is easier to prove an equivalent statement: instead of proving the negation for any [imath]A[/imath] one can show that there exists an [imath]\epsilon[/imath] such that for any [imath]\delta[/imath] a pair of points [imath]x_1,y_1[/imath] and [imath]x_2,y_2[/imath] can be found inside the [imath]\delta[/imath]-neighborhood such that [imath]|f(x_1,y_1)-f(x_2,y_2)| \geq \epsilon[/imath]. Can you see that this is an equivalent statement?

P.S. An equivalent formulation of [imath]\epsilon,\delta[/imath] definition (in [imath]L_1[/imath] metric) would use [imath]|x|+|y|<\delta[/imath] instead of [imath]x^2+y^2<\delta^2[/imath]. 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.
 
statstudent123 said:
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.
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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 [imath]\delta[/imath]-neighborhood. But I don't see any contradiction yet. The fact that [imath]f(1,3) = 1/6[/imath] does not prove anything to me. Moreover, you don't even know whether the (1,3) point is inside the [imath]\delta[/imath]-neighborhood.
On the other hand, if you can prove that in any, however small, [imath]\delta[/imath]-neighborhood one can find [imath](x,y)[/imath] such that [imath]|f(x,y)| > 2[/imath] 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 \(\displaystyle f(x,y) = \dfrac{x^2y}{2}?\)

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

​

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

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

​

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Yes sir.
 
blamocur said:
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 [imath]\delta[/imath]-neighborhood. But I don't see any contradiction yet. The fact that [imath]f(1,3) = 1/6[/imath] does not prove anything to me. Moreover, you don't even know whether the (1,3) point is inside the [imath]\delta[/imath]-neighborhood.
On the other hand, if you can prove that in any, however small, [imath]\delta[/imath]-neighborhood one can find [imath](x,y)[/imath] such that [imath]|f(x,y)| > 2[/imath] 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.
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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?
 
blamocur said:
You have to prove that the relevant [imath](x,y)[/imath] can be found for any [imath]\delta[/imath].
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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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