Multivariable calculus

Limit of (x-y)/(x+y) at the Origin Does Not Exist

The limit of (x - y)/(x + y) at the origin does not exist. Along the x-axis the function equals 1 at every point and along the y-axis it equals -1. It is also undefined on the whole line y = -x, and it grows without bound on curves that reach the origin alongside that line.

f(x,y) = \frac{x - y}{x + y}

The limit does not exist

Path	Limit along it
along y = 0	1
along x = 0	-1
along y = 2x	-0.3333333333333333

Substitute the axes and read off the constants

Both numerator and denominator are degree one, so along any line through the origin the parameter cancels completely and the function is constant.

f(t, 0) = \frac{t - 0}{t + 0} = 1, \qquad f(0, t) = \frac{0 - t}{0 + t} = -1

That is the whole proof: two paths, two different constants, no limit. The general slope confirms how badly the value varies.

f(x, mx) = \frac{x(1 - m)}{x(1 + m)} = \frac{1 - m}{1 + m}, \quad m \neq -1

With $m = 2$ the value is $-\tfrac{1}{3}$ . As $m$ approaches $-1$ the value runs off to infinity, which is a warning about the line the function is not defined on.

The excluded line matters

The denominator vanishes on all of $y = -x$ , so $f$ is undefined along an entire line through the origin, not merely at one point. Every disc around the origin, however small, meets that line.

Near that line the function is unbounded. Take the path $y = -x + x^2$ , which slides in beside the bad line.

f(t, -t + t^2) = \frac{t - (-t + t^2)}{t + (-t + t^2)} = \frac{2t - t^2}{t^2} = \frac{2 - t}{t}

That grows without bound as $t \to 0^+$ . So the function is not just path-dependent, it is unbounded in every neighbourhood of the origin. Either fact alone rules out a limit.

In polar form $f = \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}$ , with no $r$ dependence and a blow-up at $\theta = 3\pi/4$ and $\theta = -\pi/4$ , the two directions along the excluded line.

The mistake: treating 0/0 as if it forces the answer 1

Substituting the origin gives $\tfrac{0}{0}$ , and some students read the matching numerator and denominator as a sign that the answer is $1$ . The form $\tfrac{0}{0}$ carries no information about the value; it only says that direct substitution is unavailable.

A related error is to say the limit is infinite because the denominator goes to zero. Along the $x$ -axis the function is exactly $1$ the entire way in, so it is not blowing up in every direction. Both a finite-value claim and an infinite-value claim are wrong; the honest verdict is that the limit does not exist.

Check whether the function is even defined on the paths you use.
One path giving a finite value and another giving unbounded growth is already a disagreement.
State the verdict as does not exist, not as infinity, when different paths behave differently.

Frequently asked questions

Is the domain issue enough on its own to say the limit does not exist?

It depends on the convention. The strict definition asks for $f$ to be defined on a whole punctured disc around the origin, and this $f$ is not, since the line $y = -x$ is missing. Under the looser convention, which only needs the origin to be a limit point of the domain, the question is fair and the two axes still give $1$ and $-1$ .

What happens along y = -x + c for a constant c?

That line does not pass through the origin, so it is not a path into $(0,0)$ . To approach the origin while hugging the bad line you need a curve like $y = -x + x^2$ , which does pass through the origin and on which the function is unbounded.