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Math Help - Proving Multivariable Limits by the Squeeze Rule

  1. #1
    Member sinewave85's Avatar
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    Proving Multivariable Limits by the Squeeze Rule

    I am studying for a test, and I can not for the life of me figure out why the first example is correct and the second one is not.

    \lim_{(x,y)\rightarrow(0,0)}\frac{xy}{\sqrt{x^{2}+  y^{2}}} = 0

    since

    0\leq|{\frac{xy}{\sqrt{x^{2}+y^{2}}}}|\leq|{x}|

    but

    0\leq|{\frac{2x^{2}y}{x^{4}+y^{2}}}|\leq2|{y}|

    does not work since

    \lim_{(x,y)\rightarrow(0,0)}{\frac{2x^{2}y}{x^{4}+  y^{2}}}\text{ does not exist}


    I know that the second one can be disproven by paths, but I am wondering why it seems that it can be squeezed to an answer that if false. Any thoughts?
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  2. #2
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    Just out of interest, how did you get these inequalities?

    Personally, I would convert these to polars, often it makes the calculation of the limit (or the showing that it doesn't exist) much easier...
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    Member sinewave85's Avatar
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    The first inequality is an example in my text book. The second was something I tried for practice. I then realized it gave a different result than I got by paths, which worried me since I can't see why the example works and my attempt does not. My book emphasizes the squeeze rule a lot and does not really explain using polar -- could you elaborate on that a bit?
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    Member sinewave85's Avatar
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    I just realized that you might have been asking what the logic was behind the ineqality statements? The way the teacher explained it is that you can write the function as a variable times a fraction. Then you show that the fractional part has to be less than 1 but greater than 0, taking the abs value of the entire inequality, if necessary. The variable times the fraction has to be less than the variable but greater than 0. Then, since the limit of 0 is 0 and the limit of the isolated variable is 0, the limit of the function is 0.
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    Well, first it involves realising that if you are approaching the origin from any path, then the magnitude of that path is approaching 0, regardless of the path taken.

    So making \displaystyle (x, y) \to (0,0) is equivalent to making \displaystyle r \to 0, regardless of \displaystyle \theta.

    You need to remember that \displaystyle x = r\cos{\theta}, y = r\sin{\theta} and \displaystyle x^2 + y^2 = r^2.


    So in your first example...

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

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

    \displaystyle = \lim_{r \to 0}r\cos{\theta}\sin{\theta}

    \displaystyle = 0.


    Here, since you have shown that regardless of the path taken (i.e. the value of \displaystyle \theta), this function \displaystyle \to 0 as the magnitude \displaystyle r \to 0, then the limit is the same no matter what path, and the limit is \displaystyle 0.


    If you find that after you have taken the limit, you still end up with a function of \displaystyle \theta, then this limit value will change as \displaystyle \theta changes, which means that the limit does not exist.


    See what happens when you convert your second example to polars...
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    Member sinewave85's Avatar
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    Ok, that seems easier -- I got lim as r -> 0 = 2cos^(2)(omega), which as you said is not a fixed value.
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    Member sinewave85's Avatar
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    Polar definitely seems like the way to go -- much more foolproof -- but I am worried that I am going to be required to use the squeeze rule specifically on the test, ie, "Prove the limit by the Squeeze Theorem, or show that the theorem does not apply." Can you tell why why the theorem does not apply to the second example? If nothing else, it would satisfy my curiosity.
    Last edited by sinewave85; March 4th 2011 at 08:30 AM. Reason: Spelling mistake.
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