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A limit at a point of the plane is a promise about every path into it: the two-path test that refutes one, the squeeze and polar coordinates that establish one, and what continuity then asks for on top.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute the value of a function along a line or a parabola into a point, use two disagreeing paths to show that a limit does not exist, use a squeeze or polar coordinates to show that one does, and say why agreement along many paths establishes nothing on its own.
One-variable limits, where approaching a point meant from the left or from the right, and a two-sided limit existed when those two agreed. That is the idea being generalised, and the generalisation is harder than it looks: in the plane there are not two ways in but infinitely many.
A path into a point is any curve ending there; the function's values along it give a one-variable limit. The two-path test shows a limit does not exist by finding two paths whose values disagree. A function is continuous at a point when the limit there exists and equals the value of the function. A squeeze bounds $|f|$ above by something that goes to zero. In polar coordinates about the point, $r$ is the distance to it, so $r \to 0$ describes every approach at once.
Write $\lim_{(x, y) \to (a, b)} f(x, y) = L$ to mean: $f(x, y)$ can be kept as close to $L$ as required by keeping $(x, y)$ close enough to $(a, b)$. Formally, for every $\varepsilon > 0$ there is a $\delta > 0$ with
$$0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta \implies |f(x, y) - L| < \varepsilon.$$
The condition on the left is a punctured disc, not an interval, and that single difference is the whole of the new difficulty. In one variable a limit had to survive two approaches; here it has to survive every path into the point at once — straight, curved, spiralling, all of them.
The two-path test. If two paths into $(a, b)$ give different limits, then the limit does not exist. This is the only tool in the lesson that ever proves anything by computing along paths, and it proves a negative. Standard paths to try, in order:
Agreement proves nothing. Every line through the origin gives $0$ for $f = x^2y/(x^4 + y^2)$, and the parabola $y = x^2$ gives $\tfrac12$. So a table of path limits that all match is a reason to look for a proof, not a proof.
Showing a limit does exist takes an argument that covers all paths together. Two work almost always:
Squeeze. Bound $|f(x, y) - L| \le g(x, y)$ with $g \to 0$. Since $|x| \le \sqrt{x^2 + y^2}$ and $x^2 \le x^2 + y^2$, a fraction whose numerator carries more degrees than its denominator is usually squeezed in one line.
Polar coordinates. Put $x = r\cos\theta$, $y = r\sin\theta$ about the point. Every approach is now $r \to 0$, and if the expression becomes $r^n$ times something bounded independently of $\theta$, the limit is $0$. The word independently is doing the work: a bound that grows as $\theta$ varies proves nothing.
Continuity. $f$ is continuous at $(a, b)$ when the limit exists there and equals $f(a, b)$. Sums, products, quotients with non-zero denominators and compositions of continuous functions are continuous, so every polynomial is continuous everywhere and every rational function is continuous off the zero set of its denominator. That settles almost every function you will meet, and what is left — the ones defined piecewise at a single bad point — is exactly what this lesson is about.
Another way: picture
Think of the surface over a punctured disc. A one-variable limit asked whether two ends of a broken curve meet. This asks whether an entire surface closes up to a single point over the centre. A surface can perfectly well come down to $0$ along every straight groove cut through it and still ride at a different height along a curved one — which is precisely the picture of $x^2y/(x^4 + y^2)$ near the origin.
Another way: steps
Faced with a limit at a point where the formula is $0/0$:
Compare, at the origin:
$$g(x, y) = \frac{xy}{x^2 + y^2}, \qquad h(x, y) = \frac{xy}{\sqrt{x^2 + y^2}}.$$
For $g$, the line $y = mx$ gives $m/(1 + m^2)$ — a different number for every slope, so no limit. The numerator and denominator both have degree $2$, and that balance is what lets the direction survive the cancellation.
For $h$ the numerator has degree $2$ and the denominator degree $1$, so one degree is left over. In polar coordinates $h = r\cos\theta\sin\theta$, whose size is at most $r$ whatever $\theta$ is — and $r \to 0$. So the limit is $0$, proved for every path in a single line.
The degree rule of thumb: when the numerator's degree beats the denominator's, expect a limit of $0$ and prove it by squeezing; when the degrees match, expect no limit and find two paths. It is a rule of thumb rather than a theorem, because the degrees can be counted differently in $x$ and in $y$ — which is exactly the trick behind $x^2y/(x^4 + y^2)$, where $y$ has to be weighted as though it were $x^2$.
Concluding from agreement. Two paths agreeing, or twenty, is not a limit. The two-path test only ever proves a limit does not exist.
Only trying lines. $x^2y/(x^4 + y^2)$ is zero along every line and $\tfrac12$ along $y = x^2$. If the powers of $x$ and $y$ are unbalanced, try a parabola.
Polar with a $\theta$-dependent bound. Reaching $r \to 0$ is not enough if what multiplies $r^n$ can blow up as $\theta$ moves. The bound has to hold for all $\theta$ at once.
Reading existence as continuity. A limit can exist at a point where the function is undefined, or defined as something else. Continuity needs the limit and the value and their agreement. A third coordinate is not a third term bolted onto a two-dimensional formula. The cross product exists only in space, a plane is fixed by a direction rather than by a slope, and a curve in space can bend without ever leaving a surface. Checking a new formula against the plane is worth doing; assuming the plane's formula still governs is how most of the errors in this course begin.
The commonest wrong move here is to compute the limit along the axes, then along $y = x$, then along $y = 2x$, find $0$ every time, and write therefore the limit is $0$. That reasoning would be valid if the paths could be exhausted. They cannot: between any two of the paths tried there are infinitely many more, and a function can be built to ride at a different height along one of them while behaving perfectly along all the rest.
$x^2y/(x^4 + y^2)$ is that function, and it is worth memorising, because it refutes the reasoning rather than merely failing to support it: every straight line gives $0$, and the parabola $y = x^2$ gives $\tfrac12$.
So the tools have two different jobs. Paths refute: one disagreeing pair is a complete proof that no limit exists. Squeezes and polar coordinates establish: they bound the function along all paths at once. Nothing in the first family can do the second family's work, and reporting a limit from path agreement alone is claiming more than the calculation supports.
Take $\lim_{(x,y) \to (0,0)} \dfrac{x^2 - y^2}{x^2 + y^2}$. Along the $x$ axis, $y = 0$ gives $x^2/x^2 = 1$ for every $x \ne 0$.
One path, one value.
Along the $y$ axis, $x = 0$ gives $-y^2/y^2 = -1$.
A second path, a different value.
Two paths, two values, so the limit does not exist. No further work is needed and none would help: one counterexample pair is a complete proof of a negative, and the answer is not that the limit is hard to find but that there is nothing to find.
Two disagreeing paths finish the question.
Take $\lim_{(x,y) \to (0,0)} \dfrac{x^3}{x^2 + y^2}$. Every line $y = mx$ gives $x/(1 + m^2) \to 0$, so $0$ is the only candidate.
Paths suggest the value; they do not establish it.
Put $x = r\cos\theta$ and $y = r\sin\theta$. The expression becomes $r^3\cos^3\theta / r^2 = r\cos^3\theta$.
Polar coordinates describe every approach at once.
Now $|r\cos^3\theta| \le r$ for every $\theta$, and $r \to 0$, so the limit is $0$. The bound did not depend on the direction, which is the hypothesis that turns this from a family of one-variable limits into a statement about the plane.
One bound, valid in every direction, proves it.
Along $y = 0$ the function is $0$; along $x = 0$ it is $0$; along $y = mx$ it is $m^2x^3/(x^2 + m^4x^4) = m^2x/(1 + m^4x^2) \to 0$.
Every line agrees on zero.
The degrees are unbalanced — $y$ appears squared and to the fourth — so try the curve $x = y^2$: the function becomes $y^4/(y^4 + y^4) = \tfrac12$.
So the limit does not exist, even though every straight line said zero. The signal to try a curve was in the exponents: with $y^4$ sitting next to $x^2$, the natural matching path is $x$ proportional to $y^2$, and reading that off the denominator is quicker than guessing.
Put the steps of a two-path argument in order, for $f(x, y) = \dfrac{4xy}{x^2 + y^2}$ at the origin.
Number the steps in order (write the number in the box):
For $f(x, y) = \dfrac{4x^2 - 4y^2}{x^2 + y^2}$, match each approach to the origin with the value $f$ takes along it.
| $4$ | $-4$ | $0$ | No limit exists | |
|---|---|---|---|---|
| Along the $x$ axis, where $y = 0$ | ||||
| Along the $y$ axis, where $x = 0$ | ||||
| Along the line $y = x$ | ||||
| Over all paths at once, at the origin |
For $f(x, y) = \dfrac{9xy}{x^2 + y^2}$, what value does $f$ take at every point of the line $y = 3x$ other than the origin?
Answer:
For $f(x, y) = \dfrac{7x^2y}{x^4 + y^2}$, what value does $f$ take at every point of the parabola $y = x^2$ other than the origin?
Answer:
Put the steps of a two-path argument in order, for $f(x, y) = \dfrac{9xy}{x^2 + y^2}$ at the origin.
Number the steps in order (write the number in the box):
A program evaluates a function at $4$ thousand points spread along many straight paths into the origin, and every path approaches $0$. What has the program established about the limit?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $f(x, y) = \dfrac{8x^2 - 2y^2}{x^2 + y^2}$, match each approach to the origin with the value $f$ takes along it.
| $8$ | $-2$ | $3$ | No limit exists | |
|---|---|---|---|---|
| Along the $x$ axis, where $y = 0$ | ||||
| Along the $y$ axis, where $x = 0$ | ||||
| Along the line $y = x$ | ||||
| Over all paths at once, at the origin |
You can refute a limit with two paths and establish one with a bound that holds in every direction, and you can say which of those two jobs each tool does. Next: the derivatives of a function of two variables, one variable at a time.
10. Your turn: does the limit of $xy^2/(x^2 + y^4)$ exist at the origin?, step 3