Polar Coordinates for Limits at the Origin

Why Use Polar Coordinates at the Origin?

Near the origin, many expressions in two variables contain \(x^2+y^2\), or combinations of \(x\) and \(y\) that are difficult to interpret separately. Polar coordinates reorganize these expressions around distance from the origin. The radius records how close a point is to the origin, while the angle records its direction.

This separation can make a limit easier to recognize. If substitution reduces an expression to a function of the radius alone, the two-variable limit becomes a one-variable limit as the radius approaches zero. If the angle remains, however, the behavior can depend on the direction of approach, and that dependence must not be ignored.

Definition: For a point \((x,y)\ne(0,0)\), its polar coordinates satisfy \(x=r\cos\theta\) and \(y=r\sin\theta\), where \(r>0\) and \(\theta\) is an angle describing the direction from the origin. In particular, \(r=\sqrt{x^2+y^2}\).

Every nonzero point in the plane has such a representation. The angle is not unique: adding an integer multiple of \(2\pi\) describes the same direction. This does not cause a problem when studying a limit, provided we consider all directions.

Translating the Limit Condition

Suppose \(f\) is defined at points near, but not necessarily at, the origin. Its polar form is obtained by making the substitutions \(x=r\cos\theta\) and \(y=r\sin\theta\). A limit at the origin concerns every point sufficiently close to the origin, not just points on one ray.

Key criterion: The statement \(\lim_{(x,y)\to(0,0)} f(x,y)=L\) means that for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(0<\sqrt{x^2+y^2}<\delta\) implies \(|f(x,y)-L|<\varepsilon\). In polar coordinates, this requires the corresponding inequality for every angle \(\theta\) and every radius \(r\) with \(0<r<\delta\), wherever the original function is defined.

The reason for this translation is the distance identity \(r=\sqrt{x^2+y^2}\). The condition \(0<\sqrt{x^2+y^2}<\delta\) is exactly \(0<r<\delta\). And every point in that punctured disk corresponds to some such radius and angle. Thus polar coordinates change how we describe the points, but not which points the limit examines.

In particular, taking \(r\to0\) while holding one angle fixed describes just one ray. A full limit must work as \(r\to0\) across all angles. The polar form is useful partly because it makes clear whether the angle remains relevant.

When the Expression Depends Only on the Radius

If polar substitution gives an expression \(g(r)\) with no \(\theta\), then every point at a given distance from the origin has the same function value. If \(\lim_{r\to0^+}g(r)=L\), the original limit is \(L\): the one-variable estimate for \(g\) applies to every angle at once.

Worked Example: A Limit That Reduces to a Function of the Radius

Evaluate \(\lim_{(x,y)\to(0,0)}\frac{\sqrt{x^2+y^2}}{1+\sqrt{x^2+y^2}}\).

For \((x,y)\ne(0,0)\), set \(r=\sqrt{x^2+y^2}>0\). The expression becomes

\frac{\sqrt{x^2+y^2}}{1+\sqrt{x^2+y^2}} =\frac{r}{1+r}.

As \(r\to0^+\), the numerator tends to \(0\) and the denominator tends to \(1\), so the radial expression tends to \(0\). To verify the estimate directly, note that \(1+r\ge1\), so

\left|\frac{r}{1+r}-0\right| =\frac{r}{1+r} \le r.

Given \(\varepsilon>0\), choose \(\delta=\varepsilon\). If \(0<r<\delta\), then \(\left|\frac{r}{1+r}\right|\le r<\varepsilon\). This estimate holds for every angle because the expression contains no angle. Therefore the two-variable limit is \(0\). Substituting back \(r=\sqrt{x^2+y^2}\) confirms that the estimate concerns exactly the original distance to the origin.

The usefulness here is not merely that polar coordinates make the algebra shorter. They show that no direction-specific behavior remains: the value depends only on distance. This is the most direct situation for using polar coordinates in a limit.

Angular Dependence Can Prevent a Limit

After substitution, a factor involving \(\theta\) may remain. If that factor changes with direction and does not become irrelevant as \(r\) approaches zero, different paths may give different limiting values. Paths are useful for detecting this failure: if two paths approaching the origin produce different limits, the full limit does not exist.

Worked Example: An Angular Expression With No Limit

Determine whether \(\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+y^2}\) exists.

For \(r>0\), substitute \(x=r\cos\theta\) and \(y=r\sin\theta\). Then

\frac{xy}{x^2+y^2} = \frac{(r\cos\theta)(r\sin\theta)} {r^2\cos^2\theta+r^2\sin^2\theta} = \frac{r^2\cos\theta\sin\theta}{r^2(\cos^2\theta+\sin^2\theta)} = \cos\theta\sin\theta.

The factor \(r^2\) cancels, leaving an expression that depends on direction. To check whether that dependence matters, compare two paths. Along the \(x\)-axis, \(y=0\) and \(x\ne0\), so the function value is \(0\). Along the line \(y=x\), with \(x\ne0\), the value is

\frac{x\cdot x}{x^2+x^2} = \frac{x^2}{2x^2} = \frac12.

Both paths approach the origin, but their function values are respectively \(0\) and \(\frac12\). Thus the limit does not exist. This agrees with the polar expression: at \(\theta=0\), it is \(0\), while at \(\theta=\frac{\pi}{4}\), it is \(\frac12\). The two path calculations check the conclusion in the original coordinates.

This example illustrates an important point: cancelling powers of \(r\) does not automatically produce a limit. After simplification, ask whether any angular dependence remains, and whether it changes the value as the direction changes.

Using Polar Coordinates Without Losing the Domain

Polar substitution describes points other than the origin when \(r>0\). That is exactly the set of points relevant to a limit at the origin, since the limit examines nearby points with the origin removed. Still, every expression must be checked on its original domain. If a denominator vanishes at some nearby points, the function is not defined there, and the limit question concerns points in its domain.

Worked Example: Simplifying a Quotient Before Taking the Limit

Evaluate \(\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{1+x^2+y^2}\).

The denominator is positive for all real \(x\) and \(y\), so the formula is defined near the origin. In polar coordinates,

\frac{x^2+y^2}{1+x^2+y^2} = \frac{r^2\cos^2\theta+r^2\sin^2\theta} {1+r^2\cos^2\theta+r^2\sin^2\theta} = \frac{r^2}{1+r^2}.

Here we used \(\cos^2\theta+\sin^2\theta=1\). The result depends only on \(r\). Since \(1+r^2\ge1\),

\left|\frac{r^2}{1+r^2}-0\right| = \frac{r^2}{1+r^2} \le r^2.

Given \(\varepsilon>0\), take \(\delta=\sqrt{\varepsilon}\). Whenever \(0<r<\delta\), we have \(r^2<\varepsilon\), and therefore \(\left|\frac{r^2}{1+r^2}\right|<\varepsilon\). The limit is \(0\). Substituting \(r^2=x^2+y^2\) recovers the original expression and confirms that the estimate applies throughout the punctured disk.

In this example, the trigonometric terms combine exactly, leaving only radial dependence. It is the identity \(x^2+y^2=r^2\), rather than a special choice of direction, that makes the reduction possible.

Why Fixed-Angle Checks Are Not Enough

A fixed-angle calculation can test the behavior along a single ray. If two such calculations disagree, they conclusively show that a limit does not exist. But agreement along a few rays does not establish the full limit, because other directions may behave differently. A successful proof must control all points sufficiently close to the origin.

Common mistake: After writing \(\frac{xy}{x^2+y^2}=\cos\theta\sin\theta\), it is incorrect to claim the limit is \(0\) because “\(r\to0\).” The simplified expression has no \(r\) in it. Along \(\theta=0\) its value is \(0\), but along \(\theta=\frac{\pi}{4}\) its value is \(\frac12\). The correction is to retain the angular dependence and check whether it gives consistent behavior across paths.

Polar coordinates are therefore a tool for organizing a limit, not a shortcut that makes every expression a one-variable limit. They are especially effective when the algebra makes the radial dependence explicit. When angle remains, the limit question still requires attention to all directions.

Check Your Understanding

Use polar substitution to analyze the expressions, and distinguish radial behavior from behavior along individual directions.

Write \(x^2+y^2\) in polar coordinates. What does \(r\) represent geometrically?
Rewrite \(\frac{x^2+y^2}{1+x^2+y^2}\) in polar coordinates. Does the angle remain?
For \(\frac{xy}{x^2+y^2}\), calculate the values along \(y=0\) and \(y=x\). What do they imply about the limit?
Why does checking the limit along one fixed angle not, by itself, prove a two-variable limit?
If polar substitution gives \(g(r)\) with no angular dependence and \(g(r)\to L\) as \(r\to0^+\), why does that support the same limit for the original function?