# How does one show that the function $f(x,y) = \frac{x-y}{x+y}$ is discontinuous along the line $y=-x$?

How does one show that the function $f(x,y) = \left\{ \begin{array}{l l} \frac{x-y}{x + y} & \quad \text{if (x,y)$\neq$(x,-x)}\\ 1 & \quad \text{if (x,y) = (x,-x)} \end{array} \right.$

is discontinuous along the line $y=-x$? Is it sufficient to show that the limit of the function would be infinity as $y$ approaches $-x$? I'm having issues figuring out how to write this since it is not a simple point.

• It doesn't make sense to ask if a function is continuous at points where it isn't defined. Most calculus books say it's automatically discontinuous at points that aren't in the domain. I personally do not like that, but that's life. – Ted Shifrin Apr 19 '16 at 3:15
• Completely left that part out, my apologies. – BridgeSkier Apr 19 '16 at 3:18
• Ah. Well, the limit isn't infinity, but you can for sure show it isn't $1$. – Ted Shifrin Apr 19 '16 at 3:23
• Well what would it be? I guess I'm having trouble with that then. – BridgeSkier Apr 19 '16 at 3:27
• @Ted Shifrin not necessarily, some functions aren't defined at certain values yet can be made continuous by an appropriate correction, e.g. $f(x)=\frac{x}{x}$. – user238841 Apr 19 '16 at 3:28

What's the value of the function along the line $y = x$?

That would be $0/2$ except at $(0,0)$, where it's $1$.

What's the value of the function along the line $y = mx + b$?

That would be $\dfrac{x - (mx+b)}{x + (mx+b)} = \dfrac{(1-m)x + b}{(1+m)x + b}$. As $x$ approaches $\dfrac{-b}{m+1}$ this diverges to $\pm\infty$ (depending on which side you're approaching from), but then has the value $1$ at $x = \dfrac{-b}{m+1}$.

Are these continuous where they intersect the line $y = -x$?

For the first, that's at the point taking the value $1$, so it is not continuous where it meets the line $y = -x$.

For the second, that's where $-x = mx +b$, or at $x = \dfrac{-b}{m+1}$, the discontinuity.

So the function certainly has discontinuities along the line $y = -x$. It's not so hard to show that it is continuous off the line. (It is the ratio of polynomials where the denominator is nonzero.)