# Question on Showing points of discontinuities of a function are removable (or not)

The question is as follows:

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

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.

• @GitGud oh sorry, I mixed up with the notation >_< I edited my post. May 8, 2013 at 6:44
• 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) May 8, 2013 at 9:01
• 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) May 8, 2013 at 9:25

• 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? May 8, 2013 at 17:27