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Math Help - Studying the limit

  1. #1
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    Studying the limit

    I need help with this question, I am not too sure how I would go about it...
    lim
    (x,y)->(2+p2,2)
    of
    (y − 2)(x − (2 + p2))/
    x2 − 4x − y2 + 3y
    by writing f(x, y) = (2 + p2 + r cos(theta), 2 + r sin(theta)). Along which radial lines
    does the limit equal to zero?
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  2. #2
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    Re: Studying the limit

    Quote Originally Posted by calmo11 View Post
    I need help with this question, I am not too sure how I would go about it...
    lim
    (x,y)->(2+p2,2)
    of
    (y − 2)(x − (2 + p2))/
    x2 − 4x − y2 + 3y
    by writing f(x, y) = (2 + p2 + r cos(theta), 2 + r sin(theta)). Along which radial lines
    does the limit equal to zero?
    Is this what you mean?

    $\large \displaystyle{\lim_{\begin{array}{l} x \to 2+p^2 \\ y\to 2 \end{array}}}\dfrac{(y-2)(x-(2+p^2))}{x^2-4x-y^2+3y}$ ?
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  3. #3
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    Re: Studying the limit

    \large \displaystyle{\lim_{\begin{array}{l} x \to 2+sqrt{2} \\ y\to 2 \end{array}}}\dfrac{(y-2)(x-(2+sqrt{2}))}{x^2-4x-y^2+3y}
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  4. #4
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    Re: Studying the limit

    $\large \displaystyle{\lim_{\begin{array}{l} x \to 2+sqrt{2} \\ y\to 2 \end{array}}}\dfrac{(y-2)(x-(2+sqrt{2}))}{x^2-4x-y^2+3y}$
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  5. #5
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    Re: Studying the limit

    If you use \sqrt{2}, latex displays

    $\sqrt{2}$
    Thanks from topsquark
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  6. #6
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    Re: Studying the limit

    Quote Originally Posted by calmo11 View Post
    $\large \displaystyle{\lim_{\begin{array}{l} x \to 2+sqrt{2} \\ y\to 2 \end{array}}}\dfrac{(y-2)(x-(2+sqrt{2}))}{x^2-4x-y^2+3y}$
    The denominator is non-zero as you approach $(2+\sqrt 2, 2)$ so there is no problem with the limit being non-zero at that point.

    Are you sure you have the question fully correct?
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  7. #7
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    Re: Studying the limit

    The problem itself is pretty close to trivial! With the numerator being (y- 2)(x- (2+\sqrt{2})), it is obvious that the numerator is 0 at x= 2+\sqrt{2}, y= 2. It is a little harder to calculate the denominator there, but not much, and the result is NOT 0. What does that tell you?

    (Dang, typing too slow again!)
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