1. ## 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?

2. ## Re: Studying the limit

Originally Posted by calmo11
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?

$\large \displaystyle{\lim_{\begin{array}{l} x \to 2+p^2 \\ y\to 2 \end{array}}}\dfrac{(y-2)(x-(2+p^2))}{x^2-4x-y^2+3y}$ ?

3. ## 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}

4. ## Re: Studying the limit

$\large \displaystyle{\lim_{\begin{array}{l} x \to 2+sqrt{2} \\ y\to 2 \end{array}}}\dfrac{(y-2)(x-(2+sqrt{2}))}{x^2-4x-y^2+3y}$

5. ## Re: Studying the limit

If you use \sqrt{2}, latex displays

$\sqrt{2}$

6. ## Re: Studying the limit

Originally Posted by calmo11
$\large \displaystyle{\lim_{\begin{array}{l} x \to 2+sqrt{2} \\ y\to 2 \end{array}}}\dfrac{(y-2)(x-(2+sqrt{2}))}{x^2-4x-y^2+3y}$
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?

7. ## 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!)