Epsilon-delta proof for function with polar coordinates

Wingeer
Messages
76
Reaction score
0

Homework Statement


This is a subtask. I was given a function, and then asked to convert it to polar coordinates. So I did, and I also determined the limit. However they ask me to do an epsilon-delta proof.
The function is:
f(x,y)=\frac{x^6 + y^8 + x^4y^5}{x^6 + y^8}, which converted to polar coordinates should be: f(rcos\theta, rsin\theta) = 1 + \frac{(rcos\theta)^4 (rsin\theta)^5}{(rcos\theta)^6 + (rsin\theta)^8}.

Homework Equations


0 < r < \delta \to |f(rcos\theta,rsin\theta) - L| < \epsilon

The Attempt at a Solution


I thought that switching to polar coordinates and watch as r approaches zero would be enough? Is this just a straight-forward epsilon-delta proof? I could anyway need some help. I was never good at this.
 
Physics news on Phys.org
I assume you are supposed to compute the limit of f at the origin?

It is a straightforward epsilon-delta proof, but since you don't give details about what you've done so far, it's not clear what more you need to write. The basic approach you describe (watch as r approaches zero) is correct.
 
Ah, yes. I am to find the limit at the origin.

I don't know exactly where to begin. I mean, straight from the definition, I have to prove that:
0 < r < \delta \to |\frac{(rcos\theta)^4(rsin\theta)^5}{(rcos\theta)^6 + (rsin\theta)^8}| < \epsilon. What next?
 
Well, you need to find a suitable \delta which makes that implication true. If you didn't get that part, you should review how to do epsilon-delta proofs.

Now, what you need to do is make a fraction small. That usually means simultaneously making the numerator small and the denominator big (or at least not too small). Try to fill in a statement that looks like this:

Set \delta =(some expression involving \varepsilon); then when 0 < r < \delta, we have |(r \cos\theta)^4 (r \sin\theta)^5| < N and |(r \cos\theta)^6 + (r\sin\theta)^8| > D where N and D are expressions chosen so that N/D < \varepsilon. The setting for \delta is actually the last thing you figure out.
 
So, can I choose delta to be what ever I would like?
 
No; the correct choice of \delta is something that emerges at the end from your computations.

You should probably consult a tutor or a more comprehensive guide to writing epsilon-delta limit proofs.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top