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Math Help - Squeeze Theorem Help

  1. #1
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    Squeeze Theorem Help

    Hey everyone. I'm having some conflict with Squeeze Theorem for two variable functions. Here are a couple of examples:

    \lim_{(x,y)\rightarrow(0,0)} \frac{xy\cos y}{3x^2 + y^2}

    Now one can say,

    -1 \leq \frac{\cos y}{3x^2 + y^2} \leq 1

    -xy \leq \frac{xy\cos y}{3x^2 + y^2} \leq xy

    \lim_{(x,y)\rightarrow(0,0)} (-xy) = 0

    \lim_{(x,y)\rightarrow(0,0)} xy = 0

    Therefore, limit exists. Or does it?

    If we let  f(x,y) = \frac{xy\cos y}{3x^2 + y^2} and approach (0,0) along the line y = mx, this happens:

    f(x, mx) = \frac{mx^2\cos (mx)}{x^2(3 + m^2)} = \lim_{x\rightarrow0} \frac{m\cos (mx)}{3 + m^2} = \frac{m}{3 + m^2}

     \lim_{(x,y)\rightarrow(0,0)} \frac{xy\cos y}{3x^2 + y^2}

    does not exist because one gets a different answer for all non vertical line. Thus, limit does not exist. But then why did Squeeze Theorem say it did?

    Another example:

    \lim_{(x,y)\rightarrow(0,0)} \frac{x^2y^3}{2x^2+y^2}

    Now one can say,

    \frac{x^2}{2x^2+y^2} \leq 1

    Then clearly

    0 \leq \frac{x^2y^3}{2x^2+y^2} \leq |y^3|

    |y^3| \rightarrow 0 as (x,y) approaches (0,0). So by the squeeze theorem,

    |\frac{x^2y^3}{2x^2+y^2}| \rightarrow 0 as (x, y) approaches (0, 0)

    Why does the latter example hold true while the former does not? Thank you in advance!
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  2. #2
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    Quote Originally Posted by lilaziz1 View Post
    Now one can say,

    -1 \leq \frac{\cos y}{3x^2 + y^2} \leq 1
    One can say it, but it isn't true. For example, if x=1/3 and y=0 then \frac{\cos y}{3x^2 + y^2} is equal to 3.
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  3. #3
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    :O! I get it now! This is not true:  3x^2 + y^2 \geq cosy all the time. It makes sense. Thank you!
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