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Does this epsilon delta limit proof check out?

  1. Nov 11, 2012 #1
    I started learning how to do these things today and boy, they take some interesting logic. Anyway, here's my attempt at one:

    prove that the limit as (x,y) → (0,0) of [(x^2)(siny)^2]/(x^2 + 2y^2) exists

    Here's what I did:

    0<√(x^2 + y^2) < δ, |[(x^2)(siny)^2]/(x^2 + 2y^2) - 0| < ε.

    x^2 ≤ x^2 + y^2 since y^2 ≥ 0

    so it follows that
    (x^2)(siny)^2]/(x^2 + 2y^2) ≤ (siny)^2 < ε

    siny < y for all y > 0
    (siny)^2 < y^2
    (siny)^2 < x^2 + y^2 = δ^2

    (x^2)(siny)^2]/(x^2 + 2y^2) ≤ (siny)^2 < δ^2 < ε ∴ limit exists at (0,0)

    Does this work?
     
  2. jcsd
  3. Nov 11, 2012 #2
    You might have made a common mistake, but I might be wrong.

    You want to find a [itex]\delta[/itex] so that you can get from [itex]0 < x^2 + y^2 < \delta[/itex] (*) to

    [itex]| \frac{x^2 \sin^2(y)}{x^2 + 2y^2} - 0| < \epsilon[/itex] (**)

    by good old, cold, mathematical logic and stuff (it would be good to state beforehand that your trying to show that the limit is 0, which I believe it is--I mean, you did, but in English, for sake of us reading). So, it seems like you've got the right idea, but it seems you assumed both (*) and (**) and went from there, which doesn't make sense, here. Also, I don't know how you concluded that

    [itex]\sin^2(y) < \epsilon[/itex]

    because you only know that

    [itex]\frac{x^2 \sin^2(y)}{x^2 + 2y^2} < \epsilon[/itex]

    and that

    [itex]\frac{x^2 \sin^2(y)}{x^2 + 2y^2} \leq \sin^2(y)[/itex]

    (because, maybe [itex]\sin^2(y) \geq \epsilon[/itex].)

    A great trick for proving these kinds of limits is often the following. Prove that

    [itex]|\frac{x^2 \sin^2(y)}{x^2 + 2y^2}| \leq 0[/itex]

    If you can prove that, then by a theorem of limits often referred to as the "squeeze theorem," the limit must be 0. Of course, you can just write out the absolute value in its two cases and see why its true.

    EDIT: HINT: You've already done most of the work, ;)
     
    Last edited: Nov 11, 2012
  4. Nov 11, 2012 #3
    Thanks for the response!


    I tried what you said and it makes sense that the squeeze theorum would work on

    0 ≤ (x^2)(siny)^2]/(x^2 + 2y^2) ≤ (siny)^2

    as (x,y) tends to (0,0), but I'm still not sure about how I would do this with just the epsilon-delta definition. Is this a case where it's difficult to transform the epsilon inequality to resemble the delta inequality?
     
  5. Nov 13, 2012 #4
    Alright, after probably too much work, I think I've gotten it worked out with delta-epsilon proof.

    The following is the tactic I used.

    Just to give clarity, we want to show that, if, for any real number [itex]\varepsilon > 0[/itex], there exists a real number [itex]\delta > 0[/itex] such that [itex]0 < ||(x,y) - (0, 0)|| < \delta[/itex] (*), then it is true that [itex]||\frac{x^2 \sin^2(y)}{x^2 + 2y^2} - 0|| < \varepsilon[/itex] (**). Of course, we are supposing that the limit actually exists and equals [itex]0[/itex].

    Thus, we start by assuming our hypothesis (*), which can be written better as (I made a mistake in my earlier post, although I guess it might be kind of equivalent)

    [itex]0 < \sqrt{x^2 - y^2} < \delta[/itex]

    From here, we want to make our way to showing (**) is true. To do this, one way is to show that [itex]\delta[/itex] is greater than some value or expression in (**). Since I know this works (at least, I think), let us show that [itex]\delta > |x|[/itex], and then show [itex]\frac{x^2 \sin^2(y)}{x^2 + 2y^2}[/itex] is less than some expression containing [itex]|x|[/itex] (or an alternative, described in the following). That is, we first want to show

    [itex]\delta > \ldots \geq |x|[/itex]

    and then show

    [itex]\frac{x^2 \sin^2(y)}{x^2 + 2y^2} \leq \ldots \leq f(x)[/itex]

    where [itex]f(x)[/itex] is some function of [itex]x[/itex] where [itex]|x|[/itex] is in it, or when we set it to be less than [itex]\varepsilon[/itex] we can derive an expression containing [itex]|x|[/itex]. The latter is what I did.

    Well, the first thing to do with [itex]\delta[/itex] is not too difficult.

    The second thing to do with [itex]\varepsilon[/itex] is a bit difficult, as you need something specific. If you have trouble at this part, I'll give a hint. Also, note that this is all rough work. Especially the part where we let some expression be less than [itex]\varepsilon[/itex], as that is what we want to show--its that if we suppose this, then we might have an expression for [itex]\delta[/itex] that can let us actually show it.

    Honestly, if you did something along these lines in your original post, and have confidence in your answer, then maybe you have the right idea after all.
     
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