Epsilon-delta proof for function with polar coordinates

[Wingeer]

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.

 

[ystael]

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.

 

[Wingeer]

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?

 

[ystael]

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.

 

[Wingeer]

So, can I choose delta to be what ever I would like?

 

[ystael]

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.