Polar Coordinates as a Tool for Analyzing Limits of Functions of Two Variables

To evaluate the limit of a function of two variables as the point approaches the origin (0,0), namely \[ \lim_{(x,y)\to(0,0)} f(x,y), \] it is often advantageous to switch from Cartesian coordinates \((x, y)\) to polar coordinates \((\rho, \theta)\), defined by \[ x = \rho \cos\theta, \quad y = \rho \sin\theta. \]

The underlying idea is quite intuitive: if the function can be rewritten as

\[ f(x, y) = g(\rho, \theta), \]

and the limit as \( \rho \to 0 \) exists and is independent of \( \theta \), then the limit of the original function exists and equals that value:

\[ \lim_{\rho\to0^{+}} g(\rho,\theta)=L\qquad\text{for all }\theta. \]

Here, \(\rho = \sqrt{x^2 + y^2}\) denotes the radial distance from the origin, while \(\theta\) specifies the direction of approach.

Important. This technique is applicable only when the point of interest is the origin, i.e., \( \lim_{(x,y)\to(0,0)} f(x,y) \). Moreover, it is not always conclusive: the value \( L \) must be independent of \( \theta \) for the limit to exist.

Can this method be extended to limits at points other than the origin?

Yes. When the limit is taken at a point \((x_0, y_0) \neq (0,0)\), one can translate the coordinate system so that \((x_0, y_0)\) becomes the new origin. The limit can then be analyzed using polar coordinates relative to the shifted frame.

In such cases, we define shifted polar coordinates as follows: \[ x = x_0 + \rho \cos\theta, \quad y = y_0 + \rho \sin\theta. \]

This approach is referred to as the shifted polar coordinate method.

Illustrative Example

Consider the function:

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

We convert to polar coordinates by substituting \( x = \rho \cos\theta \), \( y = \rho \sin\theta \):

\[ f(x, y) = \frac{(\rho \cos\theta)^2 (\rho \sin\theta)}{(\rho \cos\theta)^2 + (\rho \sin\theta)^2} \]

\[ = \frac{\rho^3 \cos^2\theta \sin\theta}{\rho^2 (\cos^2\theta + \sin^2\theta)} \]

\[ = \rho \cos^2\theta \sin\theta \]

Since \(\cos^2\theta \sin\theta\) is bounded for all \(\theta\), and \(\rho \to 0\), it follows that the entire expression tends to 0. Therefore:

\[ \lim_{(x,y)\to(0,0)} \frac{x^2 y}{x^2 + y^2} = 0 \]

The limit exists and is independent of the direction - thus, the method is effective in this case.

Example 2

Now consider:

\[ f(x, y) = \frac{xy^2}{x^2 + y^4} \]

Using polar coordinates:

\[ f(x, y) = \frac{(\rho \cos\theta)(\rho \sin\theta)^2}{(\rho \cos\theta)^2 + (\rho \sin\theta)^4} \]

\[ = \frac{\rho^3 \cos\theta \sin^2\theta}{\rho^2 \cos^2\theta + \rho^4 \sin^4\theta} \]

\[ = \rho \cdot \frac{\cos\theta \sin^2\theta}{\cos^2\theta + \rho^2 \sin^4\theta} \]

As \(\rho \to 0\), the numerator clearly vanishes. However, the denominator tends to \(\cos^2\theta\), and may become arbitrarily small when \( \cos\theta \to 0 \), potentially causing the whole expression to diverge.

Hence, the limit depends on the direction \( \theta \), and so the two-variable limit does not exist.

Indeed, consider the curve \(x = y^2\):

\[ f(y^2, y) = \frac{y^2 \cdot y^2}{(y^2)^2 + y^4} = \frac{y^4}{2y^4} = \frac{1}{2} \]

This gives a different limit along a curved path, implying:

\[ \lim_{(x,y)\to(0,0)} \frac{xy^2}{x^2 + y^4} = \text{does not exist} \]

Therefore, using polar coordinates is not sufficient in general: one must also consider non-linear paths of approach.

Note. The limitation here lies not in polar coordinates themselves, but in the tendency to fix \(\theta\) as constant. In fact, polar coordinates allow curved trajectories via \(\theta = \theta(\rho)\). The correct criterion is that \(g(\rho,\theta)\) must converge to the same value for all such functions \(\theta(\rho)\) as \(\rho \to 0^+\).

In other words, the limit exists if and only if \(g(\rho,\theta)\) converges uniformly in \(\theta\).

Example 3

Let us now compute the limit of a function at a point other than the origin:

\[ f(x, y) = xy^2 \]

We are interested in the behavior as \((x,y) \to (1, 2)\):

\[ \lim_{(x, y) \to (1, 2)} xy^2 \]

Since the point of interest is not at the origin, we apply the shifted polar coordinate method.

We translate the origin to \((1, 2)\):

\[ x = 1 + \rho \cos\theta, \quad y = 2 + \rho \sin\theta \]

Substituting into the function yields:

\[ f(x, y) = (1 + \rho \cos\theta)(2 + \rho \sin\theta)^2 \]

Expanding the square and simplifying:

\[ = (1 + \rho \cos\theta)(4 + 4\rho \sin\theta + \rho^2 \sin^2\theta) \]

\[ = 4 + 4\rho(\cos\theta + \sin\theta) + \text{terms of order } \rho^2 \text{ and higher} \]

As \(\rho \to 0\), higher-order terms vanish, and we obtain:

\[ \lim_{\rho \to 0} f(x, y) = 4 \]

The limit is finite and independent of the direction, confirming the existence of the limit.

Note. This example illustrates how the polar coordinate method can be extended to arbitrary points in the plane by appropriately translating the coordinate system.

Further examples can be developed along the same lines.