On a line there were two ways to approach a point: from the left and from the right. Matching them was enough.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
In the plane
A point in the plane can be approached along infinitely many paths, straight or curved. Every one of them must give the same value.
The path test
Two paths that give different values prove the limit does not exist. This is a tool for disproving, not for proving.
The standard example
For xy/(x² + y²) at the origin, approaching along y = 0 gives 0 while approaching along y = x gives 1/2. No limit exists.
Why proving is harder
Checking a hundred paths proves nothing, since the hundred-and-first may differ. Proving existence needs a bound that works everywhere at once.
Continuity
f is continuous at a point when the limit exists and equals the value there. Polynomials are continuous everywhere.
Infinitely many ways to approach
In one variable a point has two sides. In the plane there are infinitely many paths in, and the limit exists only if every one gives the same value. That is a far stronger requirement.
Two paths can disprove a limit
Finding two approach paths with different limiting values shows the limit does not exist. That is the practical technique, and it is much easier than proving existence.
Proving existence is harder
Agreement along every straight line is not sufficient — a limit can fail along a parabola while holding on all lines. Establishing existence generally needs polar coordinates or a squeeze argument.
Continuity carries over
A function is continuous where its limit equals its value, as before. Polynomials in two variables are continuous everywhere, and quotients are continuous away from zeros of the denominator.
Step 2: Try It Yourself
Tap and try it out.
Contours that separate into opposing pairs mean a saddle: uphill one way, downhill the other.
Step 3: Watch an Example
One step at a time.
Watch Sana Break a Limit
Sana tests the limit of xy/(x² + y²) as (x, y) approaches the origin.
- Step 1
She first approaches along the x-axis, where y = 0, giving 0 over x², which is 0.
Step 4: Your Turn
Practice makes it stick.
The Two Paths
Problem 1 of 2
Two paths to a point give limits 3 and 7. Does the limit exist? 1 yes, 0 no.
The Polynomial
Problem 2 of 2
f(x, y) = x² + 3y is continuous everywhere. What is the limit as (x, y) approaches (2, 1)?
Test the Paths
1 of 8
How many paths must agree for a two-variable limit to exist? Enter 1 for two, 2 for four, 3 for all of them.
2 of 8
f(x, y) = 3x + y². What is the limit at (1, 2)?
3 of 8
xy/(x² + y²) along y = 0. What value does it give?
4 of 8
xy/(x² + y²) along y = x. What value does it give, as a decimal?
5 of 8
Does finding fifty agreeing paths prove a limit exists? 1 yes, 0 no.
6 of 8
f(x, y) = x² − y² . What is the limit at (3, 2)?
7 of 8
Sort each statement by whether it settles a two-variable limit.
Tap something to move it.
- Empty
- Empty
8 of 8
f is continuous at (a, b) and f(a, b) = 12. What is the limit there?
Step 5: Quick Check
Show what you know.
Question 1 of 2
Two paths give 2 and 5. Does the limit exist? 1 yes, 0 no.
Question 2 of 2
What can the path test actually prove?
What You Learned
- A point in the plane can be approached along infinitely many paths, and all must agree.
- Two paths giving different values prove the limit does not exist.
- Agreeing paths prove nothing, since another path may still differ.