4.2 Limits and Continuity

Big idea

In one variable a limit had two sides. You approached from the left, you approached from the right, and if the two agreed the limit existed. That bookkeeping was cheap because a point on a line can only be reached along a line.

In the plane the situation changes completely. A point $(a,b)$ can be approached along any of infinitely many paths: straight lines at every angle, parabolas, spirals, curves that wobble as they close in. For the limit to exist, the function values must approach the same number along every one of them. That is a far stronger demand than in one variable, and it produces a permanent asymmetry in technique. Showing a limit fails is easy, because one disagreeing pair of paths settles it. Showing a limit exists is hard, because no finite collection of paths can cover them all.

So you carry two tools. The first is the two-path test, used to prove nonexistence: find two routes into the point giving different values and stop. The second is a bounding argument, used to prove existence: trap the distance from the candidate value between zero and something that provably goes to zero. Polar coordinates are the workhorse for the second job, because $r \to 0$ captures approach from every direction at once and $|\cos\theta|$ and $|\sin\theta|$ never exceed $1$.

Continuity then reads exactly as it did before: the limit exists, the function is defined, and the two agree. What changes is where continuity can break. Instead of isolated bad points on a line, you get whole curves of trouble, such as the circle where a denominator vanishes.

Decoder

The limit of $f(x,y)$ as $(x,y)$ approaches $(a,b)$ is $L$ when the values $f(x,y)$ can be forced within any preassigned tolerance of $L$ by restricting $(x,y)$ to a small enough punctured disk around $(a,b)$.

The phrase doing the work is “punctured disk”. A disk, not an interval, so the restriction is on distance, and every direction is included automatically. Punctured, so the value at $(a,b)$ itself is irrelevant, which is why a limit can exist where the function is undefined.

The consequence people skip is the converse reading. If the limit is $L$, then every path into the point yields $L$, so two paths yielding different values contradict existence. This is the entire logic of the two-path test, and it only runs in that direction. Checking three, five or a hundred paths and finding agreement proves nothing at all. The classic trap is the function $x^2y/(x^4+y^2)$, which gives $0$ along every straight line through the origin and $1/2$ along a parabola.

To prove existence you need a statement about all points at once, not about paths. Substituting $x = r\cos\theta$ and $y = r\sin\theta$ turns “approach the origin” into “let $r \to 0$” while $\theta$ stays free. If the expression can then be bounded by a function of $r$ alone that tends to zero, no direction escapes the bound and the limit exists.

Definitions and results

Distance and disks. The open disk of radius $\delta$ about $(a,b)$ is the set of points with $\sqrt{(x-a)^2 + (y-b)^2} < \delta$. Puncturing it removes the center. Nearness in the plane always means small distance, never small in one coordinate only.

Limit. $\lim_{(x,y)\to(a,b)} f(x,y) = L$ means: for every $\varepsilon > 0$ there is a $\delta > 0$ such that $|f(x,y) - L| < \varepsilon$ for all $(x,y)$ in the punctured disk of radius $\delta$ about $(a,b)$ that lie in the domain.

Uniqueness along paths. If the limit is $L$, then for any curve approaching $(a,b)$ inside the domain, the values of $f$ along that curve approach $L$.

The two-path test. If two paths into $(a,b)$ produce different limiting values, or if one path produces no limit, then $\lim_{(x,y)\to(a,b)} f(x,y)$ does not exist. Agreement along any number of paths proves nothing.

Algebra of limits. Sums, differences, products and quotients of functions with limits behave as expected, with a nonzero denominator limit required for quotients. Polynomials in $x$ and $y$ have limits given by substitution, and so do rational functions wherever the denominator is nonzero.

Squeeze argument. If $|f(x,y) - L| \le g(r)$ for all points within some distance $r$ of $(a,b)$, where $g(r) \to 0$ as $r \to 0^+$, then the limit is $L$. In polar form centered at the origin, bounding $|f - L|$ by a multiple of a positive power of $r$ finishes the problem.

Continuity at a point. $f$ is continuous at $(a,b)$ when $(a,b)$ is in the domain, the limit there exists, and it equals $f(a,b)$. Failure of any of the three breaks continuity.

Where continuity is automatic. Polynomials are continuous everywhere. A rational function is continuous off the curve where its denominator vanishes. A composition of continuous functions is continuous, so $\sin(x^2+y)$ and $e^{xy}$ inherit continuity from their pieces.

Worked examples

A limit by substitution

Evaluate

$$ \lim_{(x,y)\to(2,-1)} \frac{x^2 y + 3x}{x + y^2} $$

The denominator at $(2,-1)$ equals $2 + 1 = 3$, which is not zero, so the function is a rational function continuous at the point and substitution is legal. The numerator is $4(-1) + 6 = 2$, so the limit is $2/3$. Nothing subtle happens when the denominator survives; save the path work for the cases where it does not.

Two lines that disagree

Show that

$$ \lim_{(x,y)\to(0,0)} \frac{xy}{x^2 + y^2} $$

does not exist. Along the $x$-axis put $y = 0$: the expression is $0/x^2 = 0$ for every $x \ne 0$, so the values approach $0$. Along the line $y = x$: the expression is $x^2/(2x^2) = 1/2$ for every $x \ne 0$, so the values approach $1/2$.

Two paths, two answers, so no single number can be the limit. In polar form the expression is $\cos\theta\sin\theta$, which depends on the direction and not on $r$ at all. The function holds a constant value along each ray out of the origin, which is why the origin can never be patched.

All lines agree and the limit still fails

Consider

$$ \lim_{(x,y)\to(0,0)} \frac{x^2 y}{x^4 + y^2} $$

Along any line $y = mx$ with $m \ne 0$, the expression becomes

$$ \frac{m x^3}{x^4 + m^2 x^2} = \frac{m x}{x^2 + m^2} $$

which tends to $0/m^2 = 0$ as $x \to 0$. Along the $x$-axis and the $y$-axis the expression is identically $0$ as well. Every straight line gives $0$.

Now approach along the parabola $y = x^2$:

$$ \frac{x^2 \cdot x^2}{x^4 + x^4} = \frac{x^4}{2x^4} = \frac{1}{2} $$

for every $x \ne 0$. A line and a parabola disagree, so the limit does not exist. Treat this as the standing warning: lines are not enough, and when the denominator mixes a fourth power with a square, try a parabola.

A limit that exists, by polar bounding

Evaluate

$$ \lim_{(x,y)\to(0,0)} \frac{x^2 y}{x^2 + y^2} $$

Substituting $x = r\cos\theta$, $y = r\sin\theta$ makes the denominator $r^2$ and the numerator $r^3\cos^2\theta\sin\theta$, so the whole expression equals $r\cos^2\theta\sin\theta$. Since $|\cos^2\theta\sin\theta| \le 1$ for every angle,

$$ \left| \frac{x^2 y}{x^2 + y^2} - 0 \right| \le r $$

and $r \to 0$ as the point approaches the origin. The bound holds for all directions simultaneously, so the limit is $0$.

Check the answer against a path or two for consistency: along $y = x$ the expression is $x^3/(2x^2) = x/2 \to 0$, and along $y = x^2$ it is $x^4/(x^2+x^4) \to 0$. These agree with $0$, as they must, though on their own they would prove nothing. Defining the function to be $0$ at the origin makes it continuous there.

Practice

These problems mix the two jobs. Some ask you to evaluate a limit, by substitution or by bounding, and others ask you to show a limit fails and to name the continuity set of a given formula.

Quiz

Five items on evaluating limits, the two-path test, polar bounding, and locating discontinuities.