Homogeneous Scaling Tests for Limits

Why Scaling Reveals the Shape of a Limit

When \((x,y)\) approaches \((0,0)\), both coordinates become small, but they need not become small at the same rate. A useful way to separate these two features is to describe a point by its distance from the origin and its direction.

Homogeneous expressions are especially suited to this separation: scaling both coordinates by the same positive factor pulls out a predictable power of that factor. The power suggests whether the expression decays toward zero, while the remaining directional factor determines whether that suggestion is justified.

Central idea: Homogeneous degree describes radial scaling, not the whole limit. To conclude that a limit exists, also control how the expression behaves as its direction varies.

Homogeneous Expressions and Their Degree

A function \(H\) is homogeneous of degree \(k\) if, whenever the expressions are defined,

H(\lambda x,\lambda y)=\lambda^k H(x,y) \qquad\text{for every }\lambda>0.

For example, \(x^2y\) is homogeneous of degree \(3\), because scaling both variables multiplies it by \(\lambda^3\). The polynomial \(x^2+y^2\) has degree \(2\). A quotient of homogeneous polynomials of degrees \(p\) and \(q\) is homogeneous of degree \(p-q\), wherever its denominator is nonzero.

Definition: The degree of a homogeneous expression records the power of the common positive scale factor that appears when every coordinate is multiplied by that factor.

To use this property near the origin, write each nonzero point as

(x,y)=(rt,rs),\qquad r=\sqrt{x^2+y^2}>0,\qquad t^2+s^2=1.

Here \(r\) is the distance to the origin, and \((t,s)\) is a unit direction. As \((x,y)\) approaches the origin, \(r\) tends to zero; the direction \((t,s)\) may vary. If \(H\) is homogeneous of degree \(k\), then

$$ H(x,y)=H(rt,rs)=r^kH(t,s). $$

The factor \(r^k\) is the radial part. The value \(H(t,s)\) is the angular, or directional, part. This decomposition is the basis of the scaling test.

When Positive Degree Guarantees Decay

Suppose \(H\) is homogeneous of degree \(k>0\), is defined at every nonzero point sufficiently near the origin, and is bounded on the unit circle. Boundedness means there is a constant \(M\) such that \(|H(t,s)|\le M\) whenever \(t^2+s^2=1\). Then

|H(x,y)| =r^k|H(t,s)| \le Mr^k.

Since \(k>0\), \(Mr^k\) tends to zero as \(r\) tends to zero. Thus \(H(x,y)\) tends to zero. The condition on the directional factor matters: positive degree by itself does not guarantee the needed bound.

Theorem: If \(H\) is homogeneous of degree \(k>0\) and bounded on the unit circle, then \(\lim_{(x,y)\to(0,0)}H(x,y)=0\), provided \(H\) is defined throughout a punctured neighborhood of the origin.

To verify the limit directly, let \(\varepsilon>0\). Choose \(\delta=(\varepsilon/(M+1))^{1/k}\). If \(0<\sqrt{x^2+y^2}<\delta\), then

|H(x,y)| \le Mr^k < M\frac{\varepsilon}{M+1} <\varepsilon.

This proves the claimed limit from the definition. If \(M=0\), the function is zero on every direction, and the conclusion holds immediately.

Worked Examples: Applying the Scaling Test

Worked Example: A Degree-One Expression That Tends to Zero

Find the limit of \(f(x,y)=\dfrac{x^2y}{x^2+y^2}\) as \((x,y)\) approaches \((0,0)\), excluding the origin where the formula is undefined.

The numerator has degree \(3\), and the denominator has degree \(2\), so the quotient has degree \(1\). Set \(x=rt\), \(y=rs\), where \(r=\sqrt{x^2+y^2}\) and \(t^2+s^2=1\). Substitution gives

f(rt,rs) =\frac{(rt)^2(rs)}{(rt)^2+(rs)^2} =\frac{r^3t^2s}{r^2(t^2+s^2)} =r\frac{t^2s}{1} =rt^2s.

Because \(t^2+s^2=1\), both \(|t|\le1\) and \(|s|\le1\). Therefore \(|t^2s|\le1\), so

|f(x,y)|=r|t^2s|\le r.

As \((x,y)\) approaches the origin, \(r\to0\), and the bound proves that \(f(x,y)\to0\). This also verifies that the angular factor stays bounded. Substituting \(y=0\), for instance, gives \(f(x,0)=0\), consistent with the result.

Worked Example: Degree Zero with Direction-Dependent Values

Determine whether \(g(x,y)=\dfrac{x^2-y^2}{x^2+y^2}\) has a limit at the origin.

Both numerator and denominator have degree \(2\), so \(g\) has degree \(0\). With \(x=rt\), \(y=rs\),

g(rt,rs) =\frac{r^2t^2-r^2s^2}{r^2t^2+r^2s^2} =\frac{t^2-s^2}{t^2+s^2} =t^2-s^2.

The radial factor has disappeared, but the directional value remains. Along the \(x\)-axis, take \((x,y)=(r,0)\), so \(t=1\), \(s=0\), and \(g(r,0)=1\). Along the \(y\)-axis, take \((x,y)=(0,r)\), so \(t=0\), \(s=1\), and \(g(0,r)=-1\). These paths approach the origin but give different values. Therefore the limit does not exist.

Direct substitution checks both values: \(g(r,0)=r^2/r^2=1\), while \(g(0,r)=-r^2/r^2=-1\), for every \(r>0\).

Worked Example: A Degree-Zero Expression with a Limit

Evaluate the limit of \(h(x,y)=\dfrac{x^4+2x^2y^2+y^4}{(x^2+y^2)^2}\) at the origin.

The numerator and denominator both have degree \(4\), so the expression has degree \(0\). Factoring the numerator gives

$$ x^4+2x^2y^2+y^4=(x^2+y^2)^2. $$

Thus \(h(x,y)=1\) for every \((x,y)\ne(0,0)\). In scaled coordinates, the same calculation is

h(rt,rs) =\frac{r^4(t^4+2t^2s^2+s^4)}{r^4(t^2+s^2)^2} =\frac{(t^2+s^2)^2}{(t^2+s^2)^2} =1.

The directional value is constant, so the limit is \(1\). Checking along any ray confirms it: for \((x,y)=(r,0)\), the numerator and denominator are both \(r^4\), giving \(h(r,0)=1\).

What Degree Zero and Negative Degree Can Tell You

The degree-zero example with different ray values illustrates an important diagnostic. If a homogeneous function of degree zero has different values in two fixed directions, it cannot have a limit at the origin: scaling along each ray leaves its value unchanged. But degree zero does not automatically imply failure, as the constant directional value in the previous example shows.

For a homogeneous function of negative degree, any direction where the function is nonzero gives another obstruction. Along that ray its magnitude scales like \(r^k\) with \(k<0\), which grows without bound as \(r\to0\). For instance, \(1/(x^2+y^2)\) has degree \(-2\) and equals \(r^{-2}\), so it cannot tend to a finite limit.

Practical guide: Positive degree suggests decay if the angular factor is bounded. Degree zero leaves the directional factor exposed. Negative degree grows along any ray with a nonzero directional value.

Why Scaling Degree Alone Can Mislead

The boundedness assumption in the theorem cannot be dropped. A directional factor may become arbitrarily large near directions where an expression is undefined. In that case, the radial factor can be cancelled by growth in the angular factor.

Worked Example: Positive Degree Without a Limit

Consider \(q(x,y)=\dfrac{y^2}{x}\), defined when \(x\ne0\). It is homogeneous of degree \(1\), because

q(\lambda x,\lambda y) =\frac{(\lambda y)^2}{\lambda x} =\lambda\frac{y^2}{x} =\lambda q(x,y).

Nevertheless, compare two paths in its domain. Along \((x,y)=(u,0)\), with \(u>0\), the value is

q(u,0)=\frac{0^2}{u}=0.

Along \((x,y)=(u^2,u)\), also with \(u>0\), both coordinates tend to zero, but

q(u^2,u)=\frac{u^2}{u^2}=1.

The two paths give different values, so the limit does not exist. The degree-one scaling identity is correct; what fails is boundedness of the directional factor. In scaled coordinates, \(q(rt,rs)=r\,s^2/t\), which can become large as \(t\) approaches zero. The path \((u^2,u)\) approaches the vertical direction, where \(t\) becomes small enough to offset the factor \(r\).

A Reliable Workflow for Homogeneous Scaling

Find the degree.
Check how the whole expression changes when \(x\) and \(y\) are both multiplied by the same positive factor.

Separate radius and direction.
Write \((x,y)=(rt,rs)\), with \(r>0\) and \(t^2+s^2=1\), then simplify using homogeneity.

Inspect the directional factor.
Determine whether it is bounded, constant, or capable of becoming large on directions relevant to the domain.

Make the conclusion at the right strength.
A uniform bound can prove decay; two distinct ray values disprove a limit. Degree alone is not a complete proof.

Common Pitfall: Treating Radial Decay as the Whole Argument

A tempting but invalid argument is: “The expression has positive degree, so its value includes a positive power of \(r\), and therefore it tends to zero.” The calculation for \(q(x,y)=y^2/x\) shows the gap. Its scaled form contains a factor \(r\), but also a factor \(s^2/t\), which is not uniformly bounded as directions approach \(t=0\).

The correction is to establish a bound that holds for every relevant direction, not just to identify the power of \(r\). For example, \( |f(x,y)|\le r\) proved the limit for the first example. When the angular factor is unbounded, a path that approaches a problematic direction may reveal failure; however, a few successful paths cannot by themselves prove a limit.

Why the Test Matters

Homogeneous scaling turns a two-variable expression near the origin into a radial power multiplied by a directional profile. That organization helps identify candidate limits and select useful estimates. It also explains why ray tests are effective for disproving a limit: along a fixed ray the direction is fixed, so only the radial power changes.

The test is not a substitute for checking all directions. A proof of a limit needs control uniform over the directions that points in the domain can take. A disproof needs only a concrete obstruction, such as two paths with different limiting values or one path along which the expression fails to approach the candidate.

Check Your Understanding

Use the scaling decomposition to answer the following questions. When a limit is claimed, identify what must be checked about the directional factor.

What is the homogeneous degree of \(\dfrac{x^3+y^3}{x^2+y^2}\), wherever it is defined?
If a degree-two homogeneous function is bounded on the unit circle, what limit does it have at the origin, and why?
Why do different values along two fixed rays disprove the existence of a limit?
Can a degree-zero homogeneous function have a limit at the origin? State the condition on its directional values that makes this possible when it is defined on every nonzero point.
What feature of \(y^2/x\) prevents its positive degree from proving decay at the origin?