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Epsilon-delta proof for function with polar coordinates

  1. Apr 20, 2010 #1
    1. The problem statement, all variables and given/known data
    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:
    [tex]f(x,y)=\frac{x^6 + y^8 + x^4y^5}{x^6 + y^8}[/tex], which converted to polar coordinates should be: [tex]f(rcos\theta, rsin\theta) = 1 + \frac{(rcos\theta)^4 (rsin\theta)^5}{(rcos\theta)^6 + (rsin\theta)^8}[/tex].

    2. Relevant equations
    [tex]0 < r < \delta \to |f(rcos\theta,rsin\theta) - L| < \epsilon[/tex]

    3. 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.
     
  2. jcsd
  3. Apr 20, 2010 #2
    I assume you are supposed to compute the limit of [tex]f[/tex] 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 [tex]r[/tex] approaches zero) is correct.
     
  4. Apr 20, 2010 #3
    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:
    [tex] 0 < r < \delta \to |\frac{(rcos\theta)^4(rsin\theta)^5}{(rcos\theta)^6 + (rsin\theta)^8}| < \epsilon[/tex]. What next?
     
  5. Apr 20, 2010 #4
    Well, you need to find a suitable [tex]\delta[/tex] 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 [tex]\delta = [/tex](some expression involving [tex]\varepsilon[/tex]); then when [tex]0 < r < \delta[/tex], we have [tex]|(r \cos\theta)^4 (r \sin\theta)^5| < N[/tex] and [tex]|(r \cos\theta)^6 + (r\sin\theta)^8| > D[/tex] where [tex]N[/tex] and [tex]D[/tex] are expressions chosen so that [tex]N/D < \varepsilon[/tex]. The setting for [tex]\delta[/tex] is actually the last thing you figure out.
     
  6. Apr 20, 2010 #5
    So, can I choose delta to be what ever I would like?
     
  7. Apr 20, 2010 #6
    No; the correct choice of [tex]\delta[/tex] 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.
     
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