# Simple limit question

#### calvino

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?

#### HallsofIvy

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 any one 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 $y= x+\epsilon$ and then letting $\epsilon$ 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).

#### calvino

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".

#### TD

Homework Helper
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!

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