Multivariable calculus

Limit of x^2/(x^2+y^2) at (0,0) Does Not Exist

Q: Is x^2/(x^2+y^2) continuous away from the origin?

Yes. For (x,y) \neq (0,0) the denominator is positive, so the quotient of two polynomials is continuous. The origin is the only point where the function has no limit, and no assigned value there can fix it.

The limit of x^2/(x^2 + y^2) at the origin does not exist. Along the x-axis the function equals 1 at every point, along the y-axis it equals 0, and along y = x it equals 1/2. In polar coordinates it is cos squared of theta, which depends only on the direction of approach.

f(x,y) = \frac{x^2}{x^2 + y^2}

The limit does not exist

Path	Limit along it
along y = 0	1
along x = 0	0
along y = x	0.5

Three paths, three answers

The numerator uses only $x$ , so travelling along the $y$ -axis kills it while the denominator survives. Travelling along the $x$ -axis makes numerator and denominator identical.

f(t, 0) = \frac{t^2}{t^2} = 1, \qquad f(0, t) = \frac{0}{t^2} = 0

Two constants, $1$ and $0$ , so the limit is already dead. The diagonal adds a third value for good measure.

f(t, t) = \frac{t^2}{t^2 + t^2} = \frac{1}{2}

In general $f(x, mx) = \frac{1}{1 + m^2}$ , which is $1$ for the flat line and shrinks toward $0$ as the line gets steep. Every value in $(0, 1]$ shows up on some line, and $0$ shows up on the $y$ -axis.

Polar form: the fraction of the squared distance

The denominator is $r^2$ , the squared distance from the origin, so the whole function measures what share of that distance comes from the $x$ direction.

f = \frac{r^2\cos^2\theta}{r^2} = \cos^2\theta

Nothing depends on $r$ . Sitting on a ray at angle $\theta$ and sliding all the way to the origin, the function never changes. Every value in $[0, 1]$ is attained on points as close to the origin as you like, so there is no candidate limit at all.

This is the cleanest diagnostic in the subject. If the polar form has no $r$ in it, the limit exists only when the $\theta$ expression is constant, and here it plainly is not.

The mistake: cancelling the y^2 because it is small

The reasoning sounds plausible: near the origin $y$ is tiny, so $x^2 + y^2 \approx x^2$ , so $f \approx 1$ . It fails because $x$ is tiny at the same time, and what matters is the ratio $y/x$ , not the size of $y$ on its own.

At the point $(10^{-6}, 10^{-6})$ both coordinates are small, and $y^2$ is exactly as large as $x^2$ , so nothing can be dropped. Approaching a point in two variables means controlling a ratio, not just shrinking each coordinate.

Dropping a term is only legal when its ratio to the kept term goes to zero along every approach.
Here $y^2/x^2$ can be anything at all near the origin.
When both terms are the same order along some path, keep them both.

Frequently asked questions

Does the limit exist if I approach along a circle around the origin?

A circle centred at the origin never gets closer to it, so it is not a valid path. Paths must have $(x,y) \to (0,0)$ . A spiral such as $r = e^{-\theta}$ with $\theta \to \infty$ does approach, and on it $f = \cos^2\theta$ keeps oscillating between $0$ and $1$ , confirming the failure.

Is x^2/(x^2+y^2) continuous away from the origin?

Yes. For $(x,y) \neq (0,0)$ the denominator is positive, so the quotient of two polynomials is continuous. The origin is the only point where the function has no limit, and no assigned value there can fix it.