Proving Non-Existence of Limit: (x^4)y/(x^2-y^2) using Prof's Hint

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In summary, the conversation revolves around proving that the limit of [(x^4)y]/[x^2-y^2] does not exist. The professor's hint is that the denominator goes to zero faster than the numerator, and the method used is to let x=(y+epsilon) and observe the function as epsilon approaches zero. However, it is mentioned that the existence of the limit depends on the (x,y) point being approached. The conversation also discusses different paths to approach the limit at (0,0) and how the method used may not directly determine the limit at that point. Finally, the professor's statement about the denominator going to zero faster than the numerator is acknowledged as a potential way to predict the existence of
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calvino
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All i need to do is show that the limit of

[(x^4)y]/[x^2-y^2] Does not exist (a proof, i guess). My prof's hint was that the denominator goes to zero faster than the numerator.

What I did was I let x=(y+epsilon), and looked at the function as epsilon goes toward zero. This leads to the denominator going to zero, and the numerator to y^5. Does this make sense? How would you do it?
 
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  • #2
Whether or not a limit exists, as well as what the limit is if it does exists, depends very strongly what (x,y) point you are converging to!

Can I assume that this limit is to be taken as (x,y)-> (0,0)?

A good way of showing a limit does not exist is to show that approaching the point along different paths gives different results or that the limit along anyone path is infinity.

What do you get if you approach (0, 0) along the path y= 0? (In other words, let y= 0 and then take x->0.) What do you get if you approach (0,0) along the line y= -x? (In other words, let y= -x and then take x->0.)

Letting [itex]y= x+\epsilon[/itex] and then letting [itex]\epsilon[/itex] go to 0 just move you to the line y= x where the function is not defined. It does not, directly, tell you anything about the limit at (0,0).
 
  • #3
Firstly,thanks for your help.

Secondly, yes, I meant the limit as (x,y)->(0,0).

I see what you mean about my method using epsilon, and I understand what you wrote about the limit differing along different paths to (0,0). One thing still bothers me, though. Why does my prof constantly mention the denominator going to 0 faster than the numerator being the key to this problem. I believe his exact words were that "x^2-y^2 goes to 0 faster than r^3".
 
  • #4
I assume your professor told you this to try and show you how it may be possible to 'predict' whether or not a limit exists. Looking if a nom/denom goes faster to 0 can be a way too see if the limit exists or not, but it isn't a solid proof!
 

FAQ: Proving Non-Existence of Limit: (x^4)y/(x^2-y^2) using Prof's Hint

1. What is the first step in proving the non-existence of a limit using the professor's hint?

The first step is to rewrite the given expression using the hint provided, which is (x^4)y/(x^2-y^2) = (x^2)(x^2)y/(x^2-y^2).

2. Why is it necessary to rewrite the expression using the hint?

Rewriting the expression allows us to break it down into two separate expressions, which makes it easier to analyze and identify any potential issues with the limit.

3. What is the second step in proving the non-existence of a limit?

The second step is to find the limits of each individual expression, (x^2) and (x^2)y/(x^2-y^2), as x approaches the desired value (e.g. infinity or a specific number).

4. How do we determine the limits of each expression?

We can use algebraic manipulation and known limits (such as the limit of x^n as x approaches infinity is infinity) to evaluate the limits of each expression.

5. What does it mean if the limits of both expressions are different?

If the limits of both expressions are different, it means that the limit of the original expression does not exist, as the two parts of the expression approach different values as x approaches the desired value.

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