Recipe

Prove a two-variable limit does not exist

Use when: you suspect \lim_{(x,y)\to(a,b)}f(x,y) fails to exist, typically because direct substitution at (a,b) gives an indeterminate form like 0/0

Goal: two different paths into (a,b) along which f settles on two different numbers, which is already a complete proof the limit does not exist

Taught in Paths and nonexistent limits

1. 1

   Substitute (a,b) directly into f. If that substitution already gives a single defined number, there is nothing for the two-path test to disprove.

   Example

   For f(x,y)=\dfrac{x^2-y^2}{x^2+y^2} at (0,0), plugging in gives \dfrac{0-0}{0+0}=\dfrac00, undefined. That is exactly when the limit is worth testing.

   Why: the two-path test exists for the case where f itself has nothing to say at the point.

2. 2

   Substitute y=b (the horizontal line through (a,b)) and take the limit as x\to a. Then substitute x=a and take the limit as y\to b. These are the cheapest paths available.

   Example

   For f(x,y)=\dfrac{x^2-y^2}{x^2+y^2} at (0,0): along y=0, f(x,0)=\dfrac{x^2}{x^2}=1 for every x\neq0, so this path gives limit 1. Along x=0, f(0,y)=\dfrac{-y^2}{y^2}=-1, so this path gives limit -1.

   • the two axis values already disagree — stop and report those two paths (skip to the last step).
   • the two axis values agree — the axes alone don't settle it; move to the next step.

   Why: the axes are the two simplest paths available, and a disagreement here is already a complete proof.

3. 3

   Substitute y=m(x-a)+b for a free parameter m, and take the limit as x\to a while treating m as a fixed but arbitrary constant.

   Example

   For f(x,y)=\dfrac{x^2y}{x^4+y^2} at (0,0): substituting y=mx gives f(x,mx)=\dfrac{mx}{m^2+x^2}. Its limit as x\to0 is 0 for every fixed m, so every straight line through the origin gives the same value.

   • the limit still depends on m after simplifying — stop and report two specific slopes with different values (skip to the last step).
   • the limit comes out the same number for every m — straight lines can't disprove it; a curved path is needed next.

   ⚡ Shortcut

   A symbolic slope m tests every straight line at once. There is no need to plug in specific numbers like m=1, m=2, m=-3 one at a time and hope one of them disagrees.

   Why: infinitely many straight lines share one algebraic shape, so one symbolic computation covers all of them at once.

4. 4

   Look at the highest power of x or y in the denominator and pick a path that makes the two sides balance. A denominator with x^4 is the clue that y=x^2 is worth trying.

   Example

   For f(x,y)=\dfrac{x^2y}{x^4+y^2}, the denominator's x^4 suggests y=x^2. Substituting: f(x,x^2)=\dfrac{x^2\cdot x^2}{x^4+x^4}=\dfrac{x^4}{2x^4}=\dfrac12 for every x\neq0. That disagrees with the 0 every straight line gave in the previous step.

   Why: a curved path can bend as it approaches, something no straight line can do, which is exactly what a denominator like x^4+y^2 is built to hide from a straight-line-only search.

5. 5

   Report both paths and their two distinct limiting values. No further paths need checking once one disagreement is in hand.

   Why: the definition of a limit requires agreement along every possible path, so exhibiting even one disagreement is already fatal, no matter how many other paths happen to agree.