Choosing a Method for a Multivariable Limit
A limit problem can look difficult because its formula hides how the variables interact. One expression may simplify after factoring; another may behave differently along different paths; a third may become easy in polar coordinates or after changing variables. The central task is to choose a method that gives enough information to justify the conclusion.
Each technique has a different role. Algebraic simplification can reveal a familiar expression on the function’s domain. A path test can show that a limit does not exist, but matching answers along a few paths does not prove that it does. Polar coordinates, a suitable coordinate change, or a bound can prove a limit when the resulting estimate controls every approach to the point.
The condition \((x,y)\in D\) matters: a formula may be undefined at the point or along a curve through it. A limit concerns the values where the function is defined, sufficiently near the point.
Algebraic Simplification: Use Identities on the Domain
Factoring and cancellation can expose the behavior of a function, but the algebraic identity must hold wherever the original function is defined. Cancellation does not change the function’s domain, and it does not permit substitution at a point where the original denominator is zero.
Worked Example: Simplifying a Quotient Near a Missing Point
Find the limit of \(f(x,y)=\frac{x^2-y^2}{x-y}\) as \((x,y)\to(1,1)\), where the formula is defined for \(x\ne y\).
Factor the numerator and use the condition \(x\ne y\), which allows cancellation:
Thus, at every point of the domain, the original expression agrees with \(x+y\). As \((x,y)\to(1,1)\), this simpler expression tends to \(1+1=2\). Therefore,
Check along the domain path \(x=1+t,\ y=1\), with \(t\ne0\). The original quotient is \(\frac{(1+t)^2-1}{t}=\frac{2t+t^2}{t}=2+t\), which tends to \(2\) as \(t\to0\). This path agrees with the simplified expression; the proof of the limit comes from the identity \(f=x+y\) on the whole domain near the point, not from this single check.
Path Tests: A Reliable Way to Disprove a Limit
If a limit exists, then the function must approach the same value along every path in its domain that tends to the point. Consequently, two paths giving different limiting values disprove the existence of the multivariable limit. The reverse inference is invalid: agreement along several selected paths does not establish what happens along all other approaches.
Worked Example: Different Paths Give Different Values
Test whether \(f(x,y)=\frac{xy}{x^2+y^2}\) has a limit as \((x,y)\to(0,0)\). The formula is defined at every point except the origin.
First take the path \(y=0\), with \(x\ne0\). Then
The limit along this path is \(0\). Now take the path \(y=x\), with \(x\ne0\). Then
The limit along this path is \(\frac12\). Since \(0\ne\frac12\), the two paths give different values. Hence the multivariable limit does not exist.
Check: Both paths approach \((0,0)\), and neither path uses the point \((0,0)\), where the formula is undefined. Thus both are valid tests within the function’s domain.
Polar Coordinates: Control Every Direction
For a limit at the origin, polar coordinates \(x=r\cos\theta,\ y=r\sin\theta\) separate distance from direction. Here \(r=\sqrt{x^2+y^2}\), so approaching the origin is equivalent to \(r\to0\). A polar formula proves a limit only if its behavior is controlled uniformly over all angles; checking a few fixed values of \(\theta\) is just another collection of path tests.
Worked Example: A Polar Bound Proves a Limit
Evaluate \(\lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}\), with the function defined away from the origin.
Substitute \(x=r\cos\theta\) and \(y=r\sin\theta\). For \(r>0\),
To bound the angular factor, use \(\cos^2\theta+\sin^2\theta=1\) and the nonnegative square \((\cos^2\theta-\sin^2\theta)^2\ge0\). Expanding the square gives
Also, expanding \((\cos^2\theta+\sin^2\theta)^2\) gives \(1=\cos^4\theta+\sin^4\theta+2\cos^2\theta\sin^2\theta\). Combining these statements yields \(1\ge4\cos^2\theta\sin^2\theta\), so \(0\le\cos^2\theta\sin^2\theta\le\frac14\). Therefore, for every angle,
The upper bound tends to \(0\) as \(r\to0\), independently of \(\theta\). By the squeeze theorem, the limit is \(0\).
Check: Along \(y=x\), the function is \(\frac{x^4}{2x^2}=\frac{x^2}{2}\) for \(x\ne0\), which tends to \(0\) and satisfies the bound. The all-angle estimate, rather than this one path, establishes the limit.
Coordinate Changes: Reveal the Structure
An invertible linear change of variables can turn complicated linear combinations into simpler coordinates. Because the change is invertible and continuous, the new coordinates tend to the transformed point exactly when the original coordinates tend to the point. This makes substitution useful for limits, provided the transformed expression is then analyzed for all approaches.
Worked Example: A Linear Change Followed by a Bound
Evaluate the limit at the origin of \(g(x,y)=\frac{(x+y)^2(x-y)^2}{(x+y)^2+(x-y)^2}\), defined away from the origin.
Set \(u=x+y\) and \(v=x-y\). This change is invertible because \(x=\frac{u+v}{2}\) and \(y=\frac{u-v}{2}\). Moreover,
Thus \((x,y)\to(0,0)\) if and only if \((u,v)\to(0,0)\). In the new coordinates, \(g=\frac{u^2v^2}{u^2+v^2}\). The square \((u^2-v^2)^2\ge0\) implies \(u^4+v^4\ge2u^2v^2\). Since \((u^2+v^2)^2=u^4+v^4+2u^2v^2\), it follows that \((u^2+v^2)^2\ge4u^2v^2\). For \((u,v)\ne(0,0)\), division by \(u^2+v^2>0\) gives
The final bound tends to \(0\) as \((x,y)\to(0,0)\), so the squeeze theorem gives a limit of \(0\).
Check: Along \(y=0\), we have \(u=v=x\), and \(g(x,0)=\frac{x^4}{2x^2}=\frac{x^2}{2}\) for \(x\ne0\). This tends to \(0\) and equals the bound in this case, consistent with the result.
Direct Bounding: When a Simple Inequality Is Enough
A coordinate change is not always necessary. If the absolute value of a function is bounded by a simple quantity that tends to zero, the squeeze theorem proves the limit. The bound must apply throughout a deleted neighborhood, not merely along selected paths.
Worked Example: Bound a Coordinate Factor by the Distance
Find the limit at the origin of \(h(x,y)=\frac{x^2y}{x^2+y^2}\), defined away from the origin.
Since \(0\le x^2\le x^2+y^2\), we have \(0\le\frac{x^2}{x^2+y^2}\le1\). Therefore,
The last expression is the distance from \((x,y)\) to the origin, so it tends to \(0\). Hence \(|h(x,y)|\to0\), and the limit of \(h\) is \(0\).
Check: Along \(y=x\), \(h(x,x)=\frac{x^3}{2x^2}=\frac{x}{2}\), which tends to \(0\); the bound gives \(|h(x,x)|\le\sqrt{2}|x|\), as required.
A Common Mistake: Treating Path Agreement as Proof
For the function \(\frac{xy}{x^2+y^2}\), one might calculate \(f(x,0)=0\) and \(f(0,y)=0\), then incorrectly conclude that the limit is \(0\). The flaw is that the two coordinate axes are only two paths among infinitely many. Along \(y=x\), the value is \(\frac12\), so the proposed conclusion is false.
The correction is to distinguish what each method establishes. A single path with behavior incompatible with a proposed limit disproves that limit. A successful proof needs control over all nearby points, supplied, for example, by an identity valid throughout the domain or by a uniform bound tending to zero.
How to Select and Justify a Technique
Before calculating, inspect the formula and the point of approach. The following sequence is a practical guide; it is not a requirement to use every method on every problem.
Identify where the formula is defined near the point and whether algebraic cancellation is valid there.
Test simple lines or curves if you suspect different directions may give different values. One disagreement proves nonexistence.
Use polar coordinates when distance and angle separate naturally, or an invertible linear change when combinations such as \(x+y\) and \(x-y\) dominate.
Show that the function is trapped between quantities with the same limit, or that its absolute value is bounded by something tending to zero.
Make clear whether you have disproved a limit by paths or proved it by an identity or an estimate valid throughout the domain near the point.
These methods matter because multivariable behavior depends on every possible approach to a point. Good problem solving is not simply choosing a convenient calculation; it is matching the calculation to a claim that controls the full domain.
Check Your Understanding
For each question, identify what the proposed technique can establish and what still needs justification.
- Why is cancellation in \(\frac{x^2-y^2}{x-y}\) valid only under the condition \(x\ne y\)?
- If two paths give different limiting values, what conclusion follows?
- Why does agreement along the two coordinate axes not prove that a multivariable limit exists?
- In a polar-coordinate proof, why must an angular bound hold uniformly for every angle?
- For an invertible linear change of variables, what must be checked to relate approach in the new coordinates to approach in the original coordinates?