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The question is as follows:

Given function: $F(x,y)=\frac{x + 2y}{sin(x+y) - cos(x-y)}$

Tasks:

a/ Find points of discontinuities

b/ Decide if the points (of discontinuities) from part a are removable

Here is my work so far:

(1) For part a, I think the points of discontinuities should have form $(0, \frac{\pi}{4} + n\pi)$ or $(\frac{\pi}{4} + n\pi, 0)$ , since they make the denominator undefined. For convenience of part b, I choose to specifically deal with the point $(0, \frac{\pi}{4})$

(2) Recall definition: A point of discontinuity $x_0$ is removable if the limits of the function under certain path are equal to each other, as they are "close" to $x_0$. In particular, if the function is 1 dimensional, we get the notion of "left" and "right" limits. But here we talk about paths of any possible direction. However, these limits are not equal to $f(x_0)$, which can be defined or undefined.

(3) I'm having trouble of "finding" such paths @_@

I come across with these two, by fix x-coordinate and vary y-coordinate: $F(x, x^2 - \frac{\pi}{4})$ and $F(x, x^2 - x - \frac{\pi}{4})$ They both have limit to be $\frac{\pi}{2\sqrt(2)}$ as x approaches 0 (by my calculation)

But what I can say about these results? I feel that discontinuities of $F(x,y)$ should be not removable, but I don't know if my thought is correct.

Would someone please help me on this question?

Thank you in advance ^^

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  • $\begingroup$ @GitGud oh sorry, I mixed up with the notation >_< I edited my post. $\endgroup$
    – Cecile
    May 8, 2013 at 6:44
  • $\begingroup$ i think set of discontinuities are (x, pi/4 +npi) and (pi/4 + npi, y) for any x,y real. ie. they are lines parellel to x and y axis ie. a grid. just use formula for sin(a) - sin(b) , and cos(b) = sin (pi/2 -b) $\endgroup$
    – rohit
    May 8, 2013 at 9:01
  • $\begingroup$ and probably it has no removable singularities,as if it is removable , Then x+2y = 0 at that point . and expression in denominator is 2 * cos(x+pi/4)*sin(y-pi/4) $\endgroup$
    – rohit
    May 8, 2013 at 9:25

1 Answer 1

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Observe that F(x,y) =(x+2y) /(2*Cos(y+pi/4)*Sin(x - pi/4)).just use formula for sin(a) - sin(b) , and cos(b) = sin (pi/2 -b)

Thus the set of discontinuities are (x, pi/4 +n*pi) and (pi/4 + n*pi, y) for any x,y real. ie. they are lines parellel to x and y axis ie. a grid.

So we have to look at point of intersection of line x+2y = 0 and the above grid.

For example at (pi/4 , -pi/8) F(x,y) =(x+2y) /(2*Cos(y+pi/4)*Sin(x - pi/4)) . Cos(-pi/8 + pi/4) is non zero. So consider Lim as (x,y)->(pi/4 , -pi/8) (x+2y) /Sin (x-pi/4) . Make substitution x' = x-pi/4 and y' = y+pi/8. we get Lim (x',y')->(0,0) x'+2y'/Sin(x'). consider curve y'=0 we get limit as 1 . while curve x'+2y' = 0 we get limit as 0.

You can try this in general on the other points, we get a expression of form Lim (x',y')->(0,0) x'+2y'/{+ or -}Sin(x') or x'+2y'/ {+ or -}Sin (y') As in both cases limit never exists it will not be a removable singularity at any point.

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  • $\begingroup$ I suddenly thought of this idea... Can I convert the original F(x,y) in terms of polar coordinates, let the radius r approaches 0? I plan to use the idea that the limit is 0 no matter what angle $\delta$ that I choose, thus the points of discontinuities are not removable. Is my idea ok? $\endgroup$
    – Cecile
    May 8, 2013 at 17:27

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