Limits of Multivariable Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 4.2: “Limits and Continuity” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/4-2-limits-and-continuity |
| Textbook used in class | Stewart, Calculus, Section 14.2: “Limits and Continuity” (Examples 1, 2, 3) |
Quick Reference
Definition: $\displaystyle\lim_{(x,y)\to(a,b)} f(x,y) = L$ means $f(x,y)$ approaches $L$ along every possible path to $(a,b)$.
Two-path test (DNE): If $f$ approaches different values along two different paths, the limit does not exist.
Polar shortcut (existence): Substitute $x = r\cos\theta$, $y = r\sin\theta$. If $|f - L| \leq g(r)$ where $g(r) \to 0$ as $r \to 0^+$, then the limit is $L$.
Motivation
In single-variable calculus, a point $x$ approaches $a$ from only two directions. In two variables, $(x,y)$ can approach $(a,b)$ along infinitely many paths: along lines of any slope, along parabolas, along spirals. The limit exists only if the function approaches the same value along all paths simultaneously.
This makes multivariable limits both harder to prove (you cannot check every path) and easier to disprove (finding just two paths that give different values is enough).
Key Concepts
1. Showing a Limit Does Not Exist
Choose two specific paths to $(a,b)$ -- typically lines $y = mx$ for different values of $m$, or curves $y = x^k$. Compute the limit along each. If the two values differ, the overall limit does not exist.
Caution: If two paths give the same value, that does NOT prove the limit exists. Finding existence requires a method that handles all paths at once.
2. Proving a Limit Exists (Polar Coordinates)
For limits as $(x,y) \to (0,0)$, substitute $x = r\cos\theta$, $y = r\sin\theta$. If the result can be bounded by a function of $r$ alone (no $\theta$ in the bound), and that function goes to 0 as $r \to 0^+$, the limit is 0.
The key fact: $|\cos\theta| \leq 1$, $|\sin\theta| \leq 1$, and $|xy| \leq (x^2+y^2)/2 = r^2/2$.
Worked Examples
Example 1. Show that $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^2 - y^2}{x^2 + y^2}$ does not exist. (Stewart 14.2, Example 1.)
Along $y = 0$: $f(x,0) = x^2/x^2 = 1 \to 1$.
Along $x = 0$: $f(0,y) = -y^2/y^2 = -1 \to -1$.
Two different values, so the limit does not exist.
Example 2. Find $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^2 y}{x^2 + y^2}$. (Stewart 14.2, Example 2.)
In polar: $f = \dfrac{r^2\cos^2\theta \cdot r\sin\theta}{r^2} = r\cos^2\theta\sin\theta$.
Then $|f| \leq r$ (since $|\cos^2\theta\sin\theta| \leq 1$), and $r \to 0$.
So the limit equals $0$.
if the limit is the same along every line $y = mx$, then the limit exists. This checks only straight-line paths. Consider $\displaystyle f(x,y) = \frac{x^2 y}{x^4 + y^2}$. Along $y = mx$: $f = \frac{mx^3}{x^4+m^2x^2} = \frac{mx}{x^2+m^2} \to 0$. Yet along $y = x^2$: $f = \frac{x^4}{2x^4} = \frac{1}{2}$. The limit does not exist. Always check curved paths or use polar coordinates.
Common Misconceptions
checking the limit along all straight lines through $(a,b)$ is sufficient to prove the limit exists.
This is the limit-as-unreachable-barrier error. Straight lines through a point are only one family of paths among infinitely many. The function $f(x,y) = x^2 y/(x^4 + y^2)$ approaches 0 along every line $y = mx$ through the origin yet approaches $1/2$ along the parabolic path $y = x^2$. A proof that the limit equals $L$ must handle all paths simultaneously, typically via an inequality bound or polar substitution.
Leveled Practice
Problem 1. Show $\displaystyle\lim_{(x,y)\to(0,0)} \frac{xy}{x^2+y^2}$ does not exist.
Show answer
Along $y = 0$: $f = 0$.
Along $y = x$: $f = x^2/(2x^2) = 1/2$.
Different values, so the limit does not exist.
Problem 2. Find $\displaystyle\lim_{(x,y)\to(0,0)}\frac{3x^2 y}{x^2+y^2}$.
Show answer
In polar: $f = 3r\cos^2\theta\sin\theta$, so $|f| \leq 3r \to 0$.
The limit is $0$.