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Path Tests for Nonexistent Limits

Use paths that approach the same point to show when a multivariable function cannot have a single limit there.

Intermediate 9 min read

What You'll Learn

  • State the path test for showing that a multivariable limit does not exist.
  • Check that each chosen path stays in the function’s domain and approaches the point in question.
  • Compute limits along parameterized paths and compare their values.
  • Use both straight-line paths and a curved path in worked examples.
  • Explain why matching limits along a few paths do not prove that a multivariable limit exists.

Why Test a Limit Along Paths?

A function of two variables can approach a point from many directions. A limit at that point exists only if the function values approach one common number as the input approaches the point, regardless of the route taken.

This makes paths useful for disproving the existence of a limit. If two valid paths approach the same point but the function approaches different values along them, there cannot be a single limit for the function. The test is powerful when it succeeds, but it has an important limitation: checking a few paths and getting the same answer does not establish that the full limit exists.

Core idea: To show that a multivariable limit does not exist, find two paths that approach the same point while the function approaches different values along those paths.

Paths and the Path Test

Suppose \(f\) is defined on a domain \(D\subseteq\mathbb{R}^2\), and consider a point \((a,b)\) that can be approached by points of \(D\) other than \((a,b)\). A path to \((a,b)\) can be described by a parameter \(t\) and coordinate functions \(x(t)\) and \(y(t)\), such that \((x(t),y(t))\) lies in the domain and approaches \((a,b)\) as \(t\to 0^+\). In symbols, we require

$$ \lim_{t\to 0^+}x(t)=a, \qquad \lim_{t\to 0^+}y(t)=b, \qquad (x(t),y(t))\in D\setminus\{(a,b)\} $$

for all sufficiently small positive \(t\). Along this path, we evaluate \(f(x(t),y(t))\) and find its one-variable limit as \(t\to 0^+\).

Theorem (Path Test for Nonexistence): If there are two paths \((x_1(t),y_1(t))\) and \((x_2(t),y_2(t))\) in the domain that both approach \((a,b)\), and the limits of \(f\) along those paths exist but are different, then \(\lim_{(x,y)\to(a,b)} f(x,y)\) does not exist. The same conclusion follows if the function fails to have a finite limit along even one valid path.

The paths need not be straight lines. Any path within the domain that approaches the point can be used. Straight lines are often a convenient first choice, but a curved path can also reveal behavior that straight-line tests miss.

Why the Path Test Works

The test follows directly from the definition of a multivariable limit. That definition requires all function values sufficiently close to the point to be close to the same proposed limit. In particular, it requires this behavior for points lying on any path in the domain.

Justification: Suppose \(\lim_{(x,y)\to(a,b)}f(x,y)=L\). For every \(\varepsilon>0\), the limit definition gives a \(\delta>0\) such that
$$ 0<\sqrt{(x-a)^2+(y-b)^2}<\delta \quad\Longrightarrow\quad |f(x,y)-L|<\varepsilon $$

Now let \((x(t),y(t))\) be any valid path approaching \((a,b)\). Since its coordinates approach \(a\) and \(b\), its distance from \((a,b)\) approaches zero. Therefore, for all sufficiently small positive \(t\), that distance is less than \(\delta\) and is not zero. The limit condition then implies

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

Because this holds for every \(\varepsilon>0\), the function values along the path must approach \(L\). Thus, if two paths give different limiting values, they cannot both have the value required by a single multivariable limit. The assumed multivariable limit is impossible.

Worked Example: Two Straight Lines Give Different Limits

Worked Example: Two Straight Lines Give Different Limits

Determine whether the following limit exists:

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

The function is defined whenever \(x^2+y^2\ne 0\), so it is defined in a punctured neighborhood of the origin. First, approach the origin along the \(x\)-axis, where \(y=0\). Parameterize this path by \((x,y)=(t,0)\), with \(t>0\). Substitution gives

$$ \frac{t\cdot 0}{t^2+0^2} = \frac{0}{t^2} = 0 \qquad (t>0). $$

Therefore the limit along the \(x\)-axis is \(0\). Now approach along the line \(y=x\), parameterized by \((x,y)=(t,t)\). Substitution gives

$$ \frac{t\cdot t}{t^2+t^2} = \frac{t^2}{2t^2} = \frac12 \qquad (t>0). $$

Thus the limit along this line is \(\frac12\). Both paths approach \((0,0)\), but the resulting limits differ. By the path test, the multivariable limit does not exist.

Check: For every \(t>0\), the first path gives function value \(0\), and the second gives \(\frac12\). These constant values have limits \(0\) and \(\frac12\), respectively, so the difference is not due to an approximation or an arithmetic error.

Worked Example: The Coordinate Axes Disagree

Worked Example: The Coordinate Axes Disagree

Test the limit

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

The denominator is positive away from the origin, so both coordinate axes, with the origin removed, lie in the domain. Along the \(x\)-axis, set \((x,y)=(t,0)\) for \(t>0\):

$$ \frac{t^2-0^2}{t^2+0^2} = \frac{t^2}{t^2} = 1. $$

The limit along this path is \(1\). Along the \(y\)-axis, set \((x,y)=(0,t)\):

$$ \frac{0^2-t^2}{0^2+t^2} = \frac{-t^2}{t^2} = -1. $$

The limit along this path is \(-1\). Since \(1\ne -1\), the multivariable limit does not exist.

Check: The two expressions simplify to constants for every \(t>0\), so their respective limits are exactly \(1\) and \(-1\).

Worked Example: A Curved Path Reveals the Difference

Worked Example: A Curved Path Reveals the Difference

Consider

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

The denominator is nonzero except at the origin. Along the \(x\)-axis, parameterize by \((x,y)=(t,0)\), where \(t>0\). Then

$$ \frac{t^2\cdot 0}{t^4+0^2} = \frac{0}{t^4} = 0. $$

The limit along the \(x\)-axis is \(0\). Now take the curved path \(y=x^2\), parameterized by \((x,y)=(t,t^2)\). Substitution gives

$$ \frac{t^2\cdot t^2}{t^4+(t^2)^2} = \frac{t^4}{t^4+t^4} = \frac{t^4}{2t^4} = \frac12 \qquad (t>0). $$

The limit along this curved path is \(\frac12\), whereas the limit along the \(x\)-axis is \(0\). Both paths approach the origin and remain in the domain for \(t>0\). Therefore, the multivariable limit does not exist.

Check: On the curved path the numerator is \(t^4\) and the denominator is \(2t^4\). Their ratio is \(\frac12\) for every \(t>0\), confirming the path limit.

How to Apply the Test Reliably

A path calculation is useful only if the path is valid and the limiting values are compared correctly. A practical procedure is:

1
Identify the point and domain.
Check where the function is defined and confirm that the chosen path can approach the target point through the domain.
2
Parameterize each path.
Write its coordinates as functions of one parameter, such as \((x,y)=(t,0)\) or \((x,y)=(t,t)\), and verify that they approach the same point.
3
Substitute into the function.
Obtain a one-variable expression and simplify it carefully, keeping the restriction that the parameter is nonzero when necessary.
4
Compute and compare the path limits.
If two values differ, the multivariable limit does not exist. State that conclusion explicitly.

In two variables, a straight line through the origin can be written as \(y=mx\), where \(m\) is its slope, or as the vertical line \(x=0\). These are possible test paths, not a complete test of existence. The next important point is to distinguish what a path comparison can prove from what it cannot.

A Common Mistake: Treating Matching Paths as Proof

Suppose someone tests the function

$$ f(x,y)=\frac{xy}{x^2+y^2} $$

along the two coordinate axes. Along \(y=0\), the value is \(0\), and along \(x=0\), the value is also \(0\). It is tempting to conclude that the limit is \(0\). That conclusion is not justified: the paths tested are only two among many possible approaches.

Correction: Test another path, such as \(y=x\). Then
$$ f(t,t)=\frac{t^2}{2t^2}=\frac12 \qquad (t>0). $$

The diagonal path gives \(\frac12\), not \(0\), so the limit does not exist. The valid inference is one-way: different path limits disprove a multivariable limit, but equal limits along a few chosen paths do not prove that it exists.

There is another basic check worth making: both paths must approach the same target point, and the function must be defined on the path points used in the calculation. A path that leaves the domain cannot provide evidence about the limit within that domain.

Why Path Tests Matter

When direct substitution produces an indeterminate expression, a path test offers a concrete way to investigate whether a single limiting value is possible. The calculations reduce to one-variable limits, which are often straightforward. A successful test can settle the question immediately by producing two incompatible values.

The test is also a useful check on intuition. A function may behave simply along coordinate axes yet vary along a diagonal or curved route. The essential question is not whether the function approaches the same value on familiar paths, but whether every valid approach must lead to that value. Path tests can conclusively rule out that possibility when they find a disagreement.

Check Your Understanding

For each question, justify your answer by describing the paths and the resulting function values or limits.

  1. State the path test for proving that a multivariable limit does not exist.
  2. For \(f(x,y)=\frac{x^2-y^2}{x^2+y^2}\), what limits are obtained along the \(x\)-axis and \(y\)-axis as \((x,y)\to(0,0)\)?
  3. Why must a path used in a test stay within the domain of the function?
  4. If three chosen paths all give the same value, does that prove the multivariable limit exists? Explain.
  5. For \(f(x,y)=\frac{x^2y}{x^4+y^2}\), calculate the function values along \((x,y)=(t,t^2)\) for \(t>0\), and state what they show when compared with the \(x\)-axis.

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