Selecting Polynomial Paths to Test Limits

Why the Choice of Path Matters

A function of two variables can approach the origin in many ways. A line \(y=mx\) tests a straight approach with slope \(m\), while a curve such as \(y=kx^2\) tests an approach whose direction changes as the point nears the origin. For rational functions, choosing a path that matches the powers in the numerator or denominator can make otherwise hidden behavior visible.

A path test is most useful when it is chosen deliberately. Substituting a path reduces the function to a one-variable expression. If two valid paths give different limiting values, then the original limit cannot exist. The challenge is to find paths that expose the function’s structure, rather than to try curves at random.

Key principle: If a limit at the origin exists, then approaching the origin along any path in the function’s domain must give that same limit. Different path limits disprove the multivariable limit.

Polynomial Paths and Their Parameters

We will focus on paths of the forms \(y=mx\) and \(y=kx^2\), where \(m\) and \(k\) are fixed real numbers. More generally, polynomial paths such as \(y=kx^p\), with a positive integer \(p\), can be useful when the powers in the formula suggest a different rate of approach.

To test a path, substitute its equation into the function and simplify for points on that path near, but not at, the origin. The path must remain in the function’s domain. In particular, if substitution makes a denominator zero at every point on the path, that path cannot be used to evaluate the function there.

Definition: A path test restricts a function \(f(x,y)\) to a curve \((x,y)=(x,g(x))\) that approaches \((0,0)\). If the resulting one-variable function has a limit different from that along another valid path, then \(\lim_{(x,y)\to(0,0)} f(x,y)\) does not exist.

Choosing a Path by Balancing Powers

Suppose important denominator terms include \(x^a\) and \(y^b\). On a path \(y=kx^p\), the term \(y^b\) becomes \(k^b x^{pb}\). A useful first choice is to set \(a=pb\), so the two terms have the same power of \(x\). Neither term then automatically overwhelms the other just because it vanishes at a different rate.

This balancing step is a guide, not a guarantee that the limit will fail. After choosing the path, substitute it into the entire function, simplify, and calculate the resulting limit. Also check the parameter values: a value that makes the leading terms cancel, or the denominator vanish along the path, may require a different path or careful domain analysis.

Worked Example: A Parabola Reveals What Lines Miss

Test whether the limit of \(f(x,y)=\frac{x^2y}{x^4+y^2}\) exists at the origin. The denominator is a sum of squares and is positive except at the origin, so the function is defined at every other point near \((0,0)\).

First take a line \(y=mx\), where \(m\) is fixed. Substitution gives

f(x,mx) =\frac{x^2(mx)}{x^4+(mx)^2} =\frac{mx^3}{x^4+m^2x^2} =\frac{mx}{x^2+m^2},\qquad x\ne 0.

If \(m\ne 0\), the numerator tends to zero and the denominator tends to \(m^2\), so the line limit is \(0\). If \(m=0\), the path is \(y=0\), and the function equals \(0\) for every \(x\ne0\). Thus every line tested gives limit \(0\).

Now balance the denominator terms \(x^4\) and \(y^2\). Setting \(2p=4\) gives \(p=2\), suggesting the parabola \(y=kx^2\). Along it, for \(x\ne0\),

f(x,kx^2) =\frac{x^2(kx^2)}{x^4+(kx^2)^2} =\frac{kx^4}{x^4+k^2x^4} =\frac{k}{1+k^2}.

The denominator \(1+k^2\) is positive for every real \(k\), so this path is valid away from the origin. With \(k=0\), the path limit is \(0\); with \(k=1\), it is \(\frac12\). These values differ, so the two-variable limit does not exist. Direct substitution into the path formula checks both values: \(0/(1+0)=0\) and \(1/(1+1)=\frac12\).

Why Different Path Limits Are Conclusive

The decisive direction is from a multivariable limit to the limits along paths. Here is the reason in terms of the limit definition. Suppose \(f(x,y)\) has limit \(L\) at the origin. For every \(\varepsilon>0\), there must be a \(\delta>0\) such that, whenever \((x,y)\) is in the domain and \(0<\sqrt{x^2+y^2}<\delta\),

|f(x,y)-L|<\varepsilon.

Any path approaching the origin eventually consists of points within this distance, except for the origin itself. Therefore the same inequality must hold along that path, which means the restricted function must tend to \(L\). If two paths instead tend to different values, no single \(L\) can satisfy this requirement. This proves that differing path limits rule out the multivariable limit.

The converse does not follow: agreement along a few paths does not ensure that all approaches give the same value. A path test is a way to disprove a limit when paths disagree, not a general proof of existence when they agree.

Using the Coefficient to Probe a Family of Curves

When a path has a parameter, its limiting value may depend on that parameter. Calculating the value as a function of the parameter can show at once whether one curve in the family behaves differently from another. The parameter choices must still give paths on which the function is defined near the origin.

Worked Example: Checking a Parabolic Family’s Domain

Consider \(g(x,y)=\frac{y-x^2}{y+x^2}\). Its domain excludes points where \(y+x^2=0\). Along a line \(y=mx\) with \(m\ne0\), substitution gives

g(x,mx) =\frac{mx-x^2}{mx+x^2} =\frac{m-x}{m+x},\qquad x\ne0,\ m+x\ne0.

For fixed nonzero \(m\), \(m+x\ne0\) for all sufficiently small \(x\), and the line limit is \(1\). Along \(y=0\), the function equals \(-1\) for \(x\ne0\). The line \(y=0\) is in the domain away from the origin because its denominator is \(x^2\). These two path limits already differ.

Now inspect the parabolas \(y=kx^2\). Substitution yields

g(x,kx^2) =\frac{kx^2-x^2}{kx^2+x^2} =\frac{k-1}{k+1},\qquad x\ne0,\ k\ne-1.

For \(k=1\), the value is \(0\); for \(k=0\), it is \(-1\). This confirms differing behavior within the parabolic family. The case \(k=-1\) must be excluded: on \(y=-x^2\), the denominator \(y+x^2\) is zero for every \(x\), so the function is undefined along that curve. The path \(k=1\) is valid since its denominator is \(2x^2\ne0\) when \(x\ne0\), and the path \(k=0\) is valid since its denominator is \(x^2\ne0\). Substitution verifies the reported values, \(0/2=0\) and \(-1/1=-1\), after the common factor \(x^2\) is removed.

When the Useful Path Is Cubic

The balance need not point to a line or a parabola. If the relevant denominator terms have powers \(x^6\) and \(y^2\), setting \(2p=6\) suggests \(y=kx^3\). Comparing lines with this cubic family illustrates how a path chosen from the degrees can reveal behavior that the straight-line tests do not show.

Worked Example: Matching a Cubic Path to the Denominator

Consider \(h(x,y)=\frac{x^3y}{x^6+y^2}\). Its denominator is positive away from the origin. Along a line \(y=mx\),

h(x,mx) =\frac{x^3(mx)}{x^6+m^2x^2} =\frac{mx^4}{x^2(x^4+m^2)} =\frac{mx^2}{x^4+m^2},\qquad x\ne0.

For \(m\ne0\), the denominator tends to \(m^2>0\), so this tends to \(0\). For \(m=0\), the function is identically zero on the path away from the origin. Thus all these line tests give \(0\).

To balance \(x^6\) and \(y^2\), choose \(y=kx^3\). For \(x\ne0\),

h(x,kx^3) =\frac{x^3(kx^3)}{x^6+(kx^3)^2} =\frac{kx^6}{x^6+k^2x^6} =\frac{k}{1+k^2}.

The denominator is positive for every real \(k\). Taking \(k=0\) gives limit \(0\), while \(k=1\) gives limit \(\frac12\). These valid paths have different limits, so the limit of \(h\) at the origin does not exist. The substitution checks the two results: \(0/(1+0)=0\) and \(1/(1+1)=\frac12\).

A Common Mistake: Treating Lines as Every Possible Approach

A frequent error is to test several lines, obtain the same value each time, and conclude that the multivariable limit exists. The example \(f(x,y)=\frac{x^2y}{x^4+y^2}\) shows why that conclusion is unjustified: every line gives \(0\), but the parabola \(y=x^2\) gives \(\frac12\).

The correction is to treat line calculations as a limited set of tests, not as a complete proof. Inspect the powers in the formula, try a polynomial path that balances important terms, and verify that the function is defined along the path. If the path limits disagree, the limit fails. If the paths tested agree, that only means these tests have not yet found a contradiction.

A Practical Path-Selection Process

Inspect the formula.
Identify the terms whose rates of vanishing or possible cancellation seem important, especially in the denominator.

Test lines first.
Substitute \(y=mx\), keeping the slope \(m\) fixed, and simplify before taking the limit.

Balance powers if needed.
For terms such as \(x^a\) and \(y^b\), try \(y=kx^p\) with \(a=pb\), then substitute into the whole function.

Check the domain and compare.
Make sure the function is defined on each path near, but not at, the origin. Compare the resulting path limits; different values disprove the multivariable limit.

Why This Method Is Useful

Polynomial paths turn a two-variable expression into a one-variable calculation while preserving the essential question: does the function have the same limiting value as the point approaches the origin? Lines provide a simple first test, and the powers in the formula suggest when a curved path may be more revealing.

The method is especially effective for disproving limits. A single pair of valid paths with distinct limiting values is conclusive. Its limitation is equally important: no finite collection of successful path tests establishes that the limit exists. For that, one needs an argument that controls approaches beyond the particular curves tested.

Check Your Understanding

For each question, show the path substitution and check that the path lies in the function’s domain near the origin.

For \(\frac{x^2y}{x^4+y^2}\), what power \(p\) balances the denominator terms on \(y=kx^p\)?
For that same function, calculate the path limit along \(y=2x^2\).
For \(\frac{y-x^2}{y+x^2}\), why is \(y=-x^2\) not a valid path for testing function values?
For \(\frac{x^3y}{x^6+y^2}\), calculate the limits along \(y=0\) and \(y=x^3\). What conclusion follows?
Why does finding the same limit along several lines not prove that the multivariable limit exists?