Multivariable calculus

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

Q: Can I use the identity to prove the limit fails in one line?

Yes. Since f = 1 + \frac{2xy}{x^2+y^2} and the quotient \frac{xy}{x^2+y^2} is known to have no limit at the origin, adding a constant and scaling cannot create one. Citing the standard example is a legitimate shortcut once you have proved it once.

The limit of (x + y)^2/(x^2 + y^2) at the origin does not exist. Along y = -x the function is 0 at every point, along y = 0 it is 1, and along y = x it is 2. In polar form it equals 1 + sin(2 theta), so the value is fixed by the direction of approach alone.

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

The limit does not exist

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

Expand first, then substitute

Expanding the numerator shows what is really going on. The squared terms match the denominator exactly, and the cross term is the troublemaker.

\frac{(x+y)^2}{x^2+y^2} = \frac{x^2 + 2xy + y^2}{x^2 + y^2} = 1 + \frac{2xy}{x^2 + y^2}

So this function is one plus twice the standard path-dependent quotient. Since $\frac{xy}{x^2+y^2}$ ranges over $[-\tfrac{1}{2}, \tfrac{1}{2}]$ on lines, $f$ ranges over $[0, 2]$ .

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

Any two of those three disagree, so the limit does not exist. The anti-diagonal $y = -x$ is the striking one: the numerator is identically zero there while the denominator is not, so the function is flat at zero all the way in.

Polar form and the range of values

Both levels are degree two, so $r$ cancels and a trigonometric identity finishes the job.

f = \frac{r^2(\cos\theta + \sin\theta)^2}{r^2} = (\cos\theta + \sin\theta)^2 = 1 + \sin 2\theta

The value runs from $0$ at $\theta = -\pi/4$ up to $2$ at $\theta = \pi/4$ , and it never depends on how close you are. Every disc around the origin contains points where $f$ is $0$ and points where $f$ is $2$ .

This also settles a question students often ask: the function is nonnegative and bounded above by $2$ , yet it still has no limit. Nice behaviour of the values says nothing about convergence.

The mistake: cancelling the square with the sum of squares

The tempting move is to write $(x+y)^2 = x^2 + y^2$ and cancel to get $1$ . That identity is false; the missing $2xy$ is precisely the term that makes the limit fail. Squaring a sum is not summing the squares.

A second version is to argue that $x + y \to 0$ and $x^2 + y^2 \to 0$ , so the answer is $\tfrac{0}{0}$ and therefore $1$ . Both statements are true and the conclusion does not follow. The form $\tfrac{0}{0}$ tells you only that direct substitution is unavailable.

Expand any binomial in the numerator before judging the size of terms.
The cross term $2xy$ is the same order as $x^2 + y^2$ , so it cannot be dropped.
Two disagreeing paths beat any amount of algebraic intuition.

Frequently asked questions

What is the limit of (x+y)^2/(x^2+y^2)^2 at the origin?

That one is unbounded rather than path-dependent. In polar form it is $\frac{1 + \sin 2\theta}{r^2}$ , which blows up along every direction except the two that run along the line $y = -x$ , namely $\theta = -\pi/4$ and $\theta = 3\pi/4$ , where $1 + \sin 2\theta$ is zero and the function is identically zero. It also has no limit, but for a different reason.

Can I use the identity to prove the limit fails in one line?

Yes. Since $f = 1 + \frac{2xy}{x^2+y^2}$ and the quotient $\frac{xy}{x^2+y^2}$ is known to have no limit at the origin, adding a constant and scaling cannot create one. Citing the standard example is a legitimate shortcut once you have proved it once.