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Math Help - Limit of functions of 2 variables

  1. #1
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    Limit of functions of 2 variables

    Find if it exists.

    After testing different paths, the limit appears to be 0, so let's try an proof. I got a bit rusty at this, I just don't know how to finish it off lol.

    We need to prove if for every there exists a such that if then

    We know that

    But

    Thus we can take but then how do we relate this back to again? I forgot lols
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  2. #2
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    You could convert to polars...

    Remember that x = r\cos{\theta}, y = r\sin{\theta}, x^2 + y^2 = r^2, then

    \lim_{(x, y) \to (0, 0)}\frac{3x^2y}{x^2 + y^2} = \lim_{r \to 0}\frac{3(r\cos{\theta})^2r\sin{\theta}}{r^2}

     = \lim_{r \to 0}\frac{3r^3\cos^2{\theta}\sin{\theta}}{r^2}

     = \lim_{r \to 0}3r\cos^2{\theta}\sin{\theta}

     = 0.
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  3. #3
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    The best way to handle a problem like this is to change to polar coordinates. That way, only the single variable, r, measure how "close" to (0, 0) you are. If the limit as r goes to 0 exists and is the same for all \theta, then the limit itself exists.

    Of course, x= r cos(\theta), y= r sin(\theta), and x^2+ y^2= r^2 so this problem becomes

    \lim_{r\to 0}\frac{3(r cos(\theta))^2(r sin(\theta))}{r^2}
    \lim_{r\to 0}3 cos^2(\theta)sin(\theta)\frac{r^3}{r^2}= 3 cos^2(\theta)sin(\theta)\lim_{r\to 0} r= 0
    for all \theta.

    If you want to do a strict " \epsilon-\delta" proof just do it as you would for one variable with r and leave \theta alone.

    (You can't take " \epsilon= \sqrt{x^2+ y^2}" for two reasons:
    1) \epsilon is given- it can be anything. It is \delta you are to "take".
    2) You must get \delta in terms of \epsilon, not x and y. It must be a constant, not a function of x and y.)
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  4. #4
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    Beat you :P
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    Usagi, you would take [LaTeX ERROR: Convert failed] and I think you'll be correct.

    Halls, Prove, here is an example from my textbook that may trouble you

    Let [LaTeX ERROR: Convert failed]

    With your method, set x = r \cos(t) and y = r \sin(t). Then the new function is

    F(r,t) = \frac{r^3\cos(t)\sin(t)}{r^2(\cos^2(t)+r^2\sin^4(t  ))} = \frac{r \cos(t)\sin(t)}{\cos^2(t)+r^2\sin^4(t)}

    Taking the limit as r approaches 0, we get 0.

    HOWEVER, let x = y^2[/MAth] in [Math]f(x,y). Now we have

    f(y^2,y) = \frac{y^4}{2y^4} = 1/2 for all values.

    Hence the limit does not exist and your method fails!
    Last edited by Vlasev; August 15th 2010 at 02:33 AM.
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  6. #6
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    Thank you all!! I understand now, thanks once again!!
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  7. #7
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    The sandwich theorem works too:

    \left| \dfrac{3x^2y}{x^2+y^2} \right| \leq \dfrac{3x^2y}{x^2} = 3 y
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  8. #8
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    Quote Originally Posted by Vlasev View Post
    Usagi, you would take [LaTeX ERROR: Convert failed] and I think you'll be correct.

    Halls, Prove, here is an example from my textbook that may trouble you

    Let [LaTeX ERROR: Convert failed]

    With your method, set x = r \cos(t) and y = r \sin(t). Then the new function is

    F(r,t) = \frac{r^3\cos(t)\sin(t)}{r^2(\cos^2(t)+r^2\sin^4(t  ))} = \frac{r \cos(t)\sin(t)}{\cos^2(t)+r^2\sin^4(t)}

    Taking the limit as r approaches 0, we get 0.

    HOWEVER, let x = y^2 in [LaTeX ERROR: Convert failed] . Now we have

    f(y^2,y) = \frac{y^4}{2y^4} = 1/2 for all values.

    Hence the limit does not exist and your method fails!

    Hi Vlasev,

    we can't use this method here, but anyway, there is no reason to apply it here, since there is not any "x+y", or "xy"...No one has said that it worked all the time, i think
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  9. #9
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    I think it is misleading to just supply a method and not mention its eventual limitation. I can see the OP trying to apply it on more tricky limits and get the wrong answer. Worse, I see the OP using it as a "proof" when in fact it is not.
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  10. #10
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    Quote Originally Posted by HallsofIvy View Post
    That way, only the single variable, r, measure how "close" to (0, 0) you are. If the limit as r goes to 0 exists and is the same for all \theta, then the limit itself exists.
    Oh, I hadn't seen this before my last post...This sentence is clearly false.
    The method that consists in changing to polar coordinates is good in a lot of cases, but theta must not be fixed.
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  11. #11
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    Quote Originally Posted by Vlasev View Post
    ... here is an example from my textbook that may trouble you

    Let [LaTeX ERROR: Convert failed]

    With your method, set x = r \cos(t) and y = r \sin(t). Then the new function is

    F(r,t) = \frac{r^3\cos(t)\sin(t)}{r^2(\cos^2(t)+r^2\sin^4(t  ))} = \frac{r \cos(t)\sin(t)}{\cos^2(t)+r^2\sin^4(t)}

    Taking the limit as r approaches 0, we get 0.

    HOWEVER, let x = y^2 in [LaTeX ERROR: Convert failed] . Now we have

    f(y^2,y) = \frac{y^4}{2y^4} = 1/2 for all values.

    Hence the limit does not exist and your method fails!
    The method works correctly if it is correctly used!... in this case the limit...

    \displaystyle \lim_{r \rightarrow 0} \frac{r\ \cos t \ \sin(t)}{\cos^{2} t+r^{2}\ \sin^4(t)}

    ... exist if and only if it is the same for any t function of r. That is verified in 'almost all' the cases and the limit is 0... but not if is...

    \displaystyle \cos t = r\ \sin^{2} t

    ... where the limit is \frac{1}{2} ...

    Kind regards

    \chi \sigma
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