Limits and continuity for several variables

We now extend the epsilon-delta idea from one variable to two variables. The key geometric change is that, in two variables, we approach a point from every direction in the plane instead of from just the left and right on a line.

Epsilon-delta definition of a limit

Let $f : \mathbb{R} \to \mathbb{R}$ and let $a \in \mathbb{R}$ . We say

\lim_{x \to a} f(x) = L

if for every $\varepsilon > 0$ there exists $\delta > 0$ such that

0 < |x-a| < \delta \quad \Longrightarrow \quad |f(x)-L| < \varepsilon.

In words: if $x$ is close enough to $a$ , then $f(x)$ is as close as we want to $L$ .

Now let $f : \mathbb{R}^2 \to \mathbb{R}$ and let $(a,b) \in \mathbb{R}^2$ . We say

\lim_{(x,y) \to (a,b)} f(x,y) = L

if for every $\varepsilon > 0$ there exists $\delta > 0$ such that

0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta \quad \Longrightarrow \quad |f(x,y)-L| < \varepsilon.

This means we look inside a punctured disk of radius $\delta$ centered at $(a,b)$ . Every point in that disk, except the center itself, must map into the $\varepsilon$ -window around $L$ .

Example 1: $z = x^2 + y^2$ and the limit at $(0,0)$

For the function

f(x,y) = x^2 + y^2,

the limit at the origin is easy to guess: the surface touches the $xy$ -plane at $(0,0,0)$ and stays nonnegative everywhere.

Epsilon-Delta Visualization: z = x² + y²

For any ε > 0, we choose δ = √ε such that:

If 0 < √(x² + y²) < δ, then |f(x,y) - 0| = x² + y² < δ² = ε

ε =0.50

δ = √ε = 0.707

ε = δ² = 0.500

Red cylinder: Error band where 0 ≤ z ≤ ε

Green region: Surface points within δ-ball

Orange circle: δ-ball boundary

The interactive visualization above shows the paraboloid with an adjustable epsilon slider. As you increase $\varepsilon$ , the delta ball (orange circle) grows, the error band (red cylinder) expands, and the highlighted green region shows all surface points within distance $\delta = \sqrt{\varepsilon}$ from the origin. Notice that the surface values within the delta ball always stay below the epsilon threshold.

We want to show that for every $\varepsilon > 0$ , there is a $\delta > 0$ such that

\sqrt{x^2+y^2} < \delta \quad \Longrightarrow \quad |x^2+y^2-0| < \varepsilon.

Since $x^2+y^2 \ge 0$ , we can rewrite the output error as

|x^2+y^2| = x^2+y^2.

If $\sqrt{x^2+y^2} < \delta$ , then

x^2+y^2 < \delta^2.

So it is enough to choose

\delta = \sqrt{\varepsilon}.

Then

0 < \sqrt{x^2+y^2} < \delta \quad \Longrightarrow \quad x^2+y^2 < \delta^2 = \varepsilon.

That is exactly the epsilon-delta condition, so the limit is $0$ .

Why checking only one coordinate is not enough

In one variable, there is only one way to approach a point from each side. In two variables, there are infinitely many paths. Fixing $x=0$ or $y=0$ only checks two of them.

Remark. To prove a limit does not exist, it is enough to find two paths approaching the same point that give different limits. But if two paths give the same limit, that does not prove the limit exists.

Example 2: $z = \dfrac{x^2-y^2}{x^2+y^2}$

Consider

f(x,y) = \frac{x^2-y^2}{x^2+y^2}, \qquad (x,y) \ne (0,0).

This function looks innocent, but it behaves differently along different paths to the origin.

Example 2: Surface with three distinct approach paths

This surface shows (x² - y²)/(x² + y²). Notice the three colored paths approach different limits: blue (x-axis) → 1, green (y-axis) → −1, orange (diagonal) → 0.

The 3D visualization above shows the surface with three colored paths marked on it:

Blue path (x-axis, y=0): approaches limit value 1
Green path (y-axis, x=0): approaches limit value −1
Orange path (diagonal, y=x): approaches limit value 0

The xy-plane plot below shows where these paths are located in the domain:

xy-plane: Domain showing where the three paths lie

Along these paths, the function values are:

f(x,0) = \frac{x^2}{x^2} = 1, \qquad f(0,y) = \frac{-y^2}{y^2} = -1, \qquad f(t,t) = \frac{t^2-t^2}{t^2+t^2} = 0.

So the function approaches three different values. Therefore the limit at $(0,0)$ does not exist.

Suppose the limit existed and equal to some number $L$ .

Then every path to $(0,0)$ would have to produce the same limit $L$ .

But along the x-axis the limit is $1$ , along the y-axis it is $-1$ , and along the diagonal it is $0$ .

Since a single limit cannot equal three different numbers, the limit does not exist.

Example 3: $z = \dfrac{xy}{x^2+y^2}$

Now look at

g(x,y) = \frac{xy}{x^2+y^2}, \qquad (x,y) \ne (0,0).

Again, different paths give different answers.

Example 3: Surface with three distinct approach paths

This surface shows xy/(x² + y²). Notice the three colored paths: blue and green (x and y axes) → 0, but orange (diagonal) → 1/2.

The 3D visualization shows the surface with three colored paths:

Blue path (x-axis, y=0): approaches limit value 0
Green path (y-axis, x=0): approaches limit value 0
Orange path (diagonal, y=x): approaches limit value 1/2

The xy-plane diagram below shows where these paths are located in the domain:

xy-plane: Domain showing where the three paths lie

Along the x-axis,

g(x,0) = 0.

Along the y-axis,

g(0,y) = 0.

Along the line $y=x$ ,

g(t,t) = \frac{t^2}{2t^2} = \frac12.

Therefore the limit at $(0,0)$ does not exist, because one path gives $0$ and another gives $\frac12$ .

Example 4: $z = \dfrac{3x^2y}{x^2+y^2}$ has limit $0$

Now consider

h(x,y) = \frac{3x^2y}{x^2+y^2}, \qquad (x,y) \ne (0,0).

This example is useful because the limit does exist, but the algebra is slightly more subtle than in Example 1.

Example 4: All paths converge to the same limit

This surface shows 3x²y/(x² + y²). All three colored paths converge to 0: blue (x-axis) → 0, green (y-axis) → 0, orange (diagonal) → 0. This is why the limit exists!

Notice in the visualization that all three colored paths approach the same value of $0$ . This is consistent with the limit existing! Unlike Examples 2 and 3, here every approach path (and in fact, every approach path) gives the same limiting value.

We estimate the absolute value:

\left|\frac{3x^2y}{x^2+y^2}\right| = 3|y|\frac{x^2}{x^2+y^2}.

Since

0 \le \frac{x^2}{x^2+y^2} \le 1,

we get

\left|\frac{3x^2y}{x^2+y^2}\right| \le 3|y|.

Now if $(x,y) \to (0,0)$ , then in particular $y \to 0$ , so $3|y| \to 0$ . By the squeeze theorem, the whole expression goes to $0$ .

If you want a version written directly in $\varepsilon$ - $\delta$ form, choose $\delta = \varepsilon/3$ . Then

\sqrt{x^2+y^2} < \delta \Longrightarrow |y| < \delta \Longrightarrow \left|\frac{3x^2y}{x^2+y^2}\right| < 3\delta = \varepsilon.

So the limit is $0$ .

Continuity

Let $f : \mathbb{R}^2 \to \mathbb{R}$ and let $(a,b)$ be in the domain of $f$ .

We say that $f$ is continuous at $(a,b)$ if

\lim_{(x,y) \to (a,b)} f(x,y) = f(a,b).

Equivalently, for every $\varepsilon > 0$ there exists $\delta > 0$ such that

\sqrt{(x-a)^2 + (y-b)^2} < \delta \quad \Longrightarrow \quad |f(x,y)-f(a,b)| < \varepsilon.

If a function is built from sums, products, quotients, and compositions of continuous functions, then it is continuous wherever it is defined.

Exercises

State the definition of $\lim_{(x,y)\to(a,b)} f(x,y)=L$ using the phrase "inside a punctured disk."

Answer: For every $\varepsilon > 0$ , there exists $\delta > 0$ such that whenever $0 < \sqrt{(x-a)^2+(y-b)^2} < \delta$ , we have $|f(x,y)-L| < \varepsilon$ .

Investigate

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

at $(0,0)$ .

Answer: Along the x-axis and y-axis the value is $0$ . Along $y=x$ the value is $\frac{x}{2}$ , which tends to $0$ as well. This does not prove the limit exists, but in fact the estimate

\left|\frac{x^2y}{x^2+y^2}\right| \le |y|

shows the limit is $0$ .

Show that

\frac{x^2-y^2}{x^2+y^2}

has no limit at $(0,0)$ by checking the path $y=mx$ .

Answer: Along $y=mx$ ,

\frac{x^2-m^2x^2}{x^2+m^2x^2} = \frac{1-m^2}{1+m^2}.

This depends on $m$ , so different slopes give different limits. Therefore the limit does not exist.

f(x,y) = \frac{x^2+y^2}{1+x^2+y^2}

continuous at $(0,0)$ ?

Answer: Yes. The denominator is never $0$ , so the function is defined everywhere. Since it is built from polynomials and a quotient with nonzero denominator, it is continuous everywhere. In particular,

f(0,0) = 0,

and the limit at $(0,0)$ is also $0$ .

Limits and Continuity for Several Variables

Limits and continuity for several variables

Epsilon-delta definition of a limit

Example 1: z=x2+y2z = x^2 + y^2z=x2+y2 and the limit at (0,0)(0,0)(0,0)

Epsilon-Delta Visualization: z = x² + y²

Why checking only one coordinate is not enough

Example 2: z=x2−y2x2+y2z = \dfrac{x^2-y^2}{x^2+y^2}z=x2+y2x2−y2​

Example 2: Surface with three distinct approach paths

xy-plane: Domain showing where the three paths lie

Example 3: z=xyx2+y2z = \dfrac{xy}{x^2+y^2}z=x2+y2xy​

Example 3: Surface with three distinct approach paths

xy-plane: Domain showing where the three paths lie

Example 4: z=3x2yx2+y2z = \dfrac{3x^2y}{x^2+y^2}z=x2+y23x2y​ has limit 000

Example 4: All paths converge to the same limit

Continuity

Exercises

Example 1: $z = x^2 + y^2$ and the limit at $(0,0)$

Example 2: $z = \dfrac{x^2-y^2}{x^2+y^2}$

Example 3: $z = \dfrac{xy}{x^2+y^2}$

Example 4: $z = \dfrac{3x^2y}{x^2+y^2}$ has limit $0$