Limits and Continuity of Multivariable Functions

Welcome to your lesson on limits and continuity in multivariable calculus! 🎉 Today, you’ll dive into the fascinating world of limits and continuity for functions of several variables. By the end of this lesson, you’ll be able to confidently evaluate limits, understand what it means for a function to be continuous in two or more dimensions, and avoid common pitfalls along the way. Let’s get started and unlock the keys to understanding how functions behave in higher dimensions!

Understanding Limits in Multivariable Functions

Before we jump into multivariable limits, let’s recall the single-variable case. In single-variable calculus, the limit of a function $f(x)$ as $x$ approaches a value $a$ is the value that $f(x)$ gets closer to as $x$ gets closer to $a$. Simple, right? Now, let’s extend this idea to two or more variables.

Definition of a Limit in Two Variables

Let’s consider a function $f(x, y)$ of two variables. We say that the limit of $f(x, y)$ as $(x, y)$ approaches $(a, b)$ is $L$ if:

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

This means that as the point $(x, y)$ gets arbitrarily close to $(a, b)$, the value of $f(x, y)$ gets arbitrarily close to $L$.

But here’s the tricky part: in two dimensions, we can approach the point $(a, b)$ from infinitely many paths—along the $x$-axis, the $y$-axis, diagonally, or even along curves. For the limit to exist, the value of $f(x, y)$ must approach the same number $L$ regardless of the path we take.

Example: A Simple Multivariable Limit

Let’s look at a simple example to make this clearer. Consider the function:

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

We want to find:

$\lim_{(x,y) \to (0,0)} f(x,y)$

We can approach $(0,0)$ along different paths. Let’s try a few:

Along the $x$-axis: Set $y = 0$. Then:

f(x, 0) = $\frac{x^2 \cdot 0}{x^2 + 0}$ = 0

As $x \to 0$, $f(x, 0) \to 0$.

Along the $y$-axis: Set $x = 0$. Then:

f(0, y) = $\frac{0 \cdot y}{0 + y^2}$ = 0

As $y \to 0$, $f(0, y) \to 0$.

Along the line $y = x$: Substitute $y = x$. Then:

f(x, x) = $\frac{x^2 \cdot x}{x^2 + x^2}$ = $\frac{x^3}{2x^2}$ = $\frac{x}{2}$

As $x \to 0$, $f(x, x) \to 0$.

In all these cases, the limit is $0$. It seems like the limit is $0$ regardless of the path we take. To be rigorous, we can use polar coordinates.

Using Polar Coordinates

One of the most powerful tools for evaluating limits in two dimensions is polar coordinates. In polar coordinates, we let:

x = r $\cos$ $\theta$, \quad y = r $\sin$ $\theta$

Then $r$ is the distance from the origin, and $\theta$ is the angle.

Let’s rewrite $f(x, y)$ in polar coordinates:

f(r, $\theta)$ = $\frac{(r \cos \theta)^2 (r \sin \theta)}{(r \cos \theta)^2 + (r \sin \theta)^2}$ = $\frac{r^3 \cos^2 \theta \sin \theta}{r^2 (\cos^2 \theta + \sin^2 \theta)}$

Since $\cos^2 \theta + \sin^2 \theta = 1$, this simplifies to:

f(r, $\theta)$ = r $\cos^2$ $\theta$ $\sin$ $\theta$

Now, as $r \to 0$, $f(r, \theta) \to 0$ regardless of $\theta$. So the limit is $0$.

Common Pitfall: Different Limits Along Different Paths

Sometimes, the limit does not exist. This happens when the limit depends on the path. Here’s a classic example:

g(x, y) = $\frac{x^2 - y^2}{x^2 + y^2}$

Let’s find:

$\lim_{(x,y) \to (0,0)} g(x,y)$

Along the line $y = 0$:

g(x, 0) = $\frac{x^2 - 0}{x^2 + 0}$ = 1

As $x \to 0$, $g(x, 0) \to 1$.

Along the line $x = 0$:

g(0, y) = $\frac{0 - y^2}{0 + y^2}$ = -1

As $y \to 0$, $g(0, y) \to -1$.

We’ve found two different values depending on the path: along the $x$-axis, the limit is $1$, and along the $y$-axis, the limit is $-1$. Because the limit depends on the path, the limit does not exist.

Key Takeaways

For a limit to exist in multiple dimensions, the function must approach the same value along every possible path.
Polar coordinates can simplify the process of evaluating limits.
If different paths yield different values, the limit does not exist.

Continuity in Multivariable Functions

Now that we understand limits, let’s talk about continuity. In single-variable calculus, a function $f(x)$ is continuous at $x = a$ if:

$f(a)$ is defined.
$\lim_{x \to a} f(x)$ exists.
$\lim_{x \to a} f(x) = f(a)$.

We extend this to multivariable functions.

Definition of Continuity

A function $f(x, y)$ is continuous at a point $(a, b)$ if:

$f(a, b)$ is defined.
$\lim_{(x,y) \to (a,b)} f(x,y)$ exists.
$\lim_{(x,y) \to (a,b)} f(x,y) = f(a, b)$.

If a function is continuous at every point $(x, y)$ in its domain, we say the function is continuous on its domain.

Example: A Continuous Function

Consider the function:

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

Let’s check if it’s continuous at $(0,0)$.

Is $f(0,0)$ defined? Yes, $f(0,0) = 0^2 + 0^2 = 0$.
Does the limit exist? Let’s use polar coordinates:

f(r, $\theta)$ = (r $\cos$ $\theta)^2$ + (r $\sin$ $\theta)^2$ = r^2 ($\cos^2$ $\theta$ + $\sin^2$ $\theta)$ = r^2

As $r \to 0$, $f(r, \theta) \to 0$ regardless of $\theta$.

So, the limit exists and is $0$.

Is the limit equal to $f(0,0)$? Yes, the limit is $0$ and $f(0,0) = 0$.

So, $f(x, y)$ is continuous at $(0,0)$. In fact, this function is continuous everywhere because it’s a polynomial in $x$ and $y$, and polynomials are continuous everywhere.

Example: A Function That’s Not Continuous

Let’s revisit the function:

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

We found earlier that:

$\lim_{(x,y) \to (0,0)} f(x,y) = 0$

But is $f(x, y)$ continuous at $(0,0)$?

Is $f(0,0)$ defined? Let’s check:

f(0,0) = $\frac{0^2 \cdot 0}{0^2 + 0^2}$ = $\frac{0}{0}$

Oops! $f(0,0)$ is not defined. So, $f(x, y)$ is not continuous at $(0,0)$.

We can fix this by defining $f(0,0) = 0$. With this new definition, $f(x, y)$ becomes continuous at $(0,0)$.

Piecewise Functions and Continuity

Sometimes functions are defined piecewise. Let’s consider:

$h(x, y) = \begin{cases}$

$\frac{xy}{x^2 + y^2}$ & \text{if } (x, y) $\neq$ (0, 0) \\

0 & \text{if } (x, y) = (0, 0)

$\end{cases}$

Is $h(x, y)$ continuous at $(0,0)$?

Is $h(0,0)$ defined? Yes, $h(0,0) = 0$.
Does the limit exist? We found earlier that the limit of $\frac{xy}{x^2 + y^2}$ as $(x, y) \to (0,0)$ is $0$.
Is the limit equal to $h(0,0)$? Yes, both are $0$.

So, $h(x, y)$ is continuous at $(0,0)$.

Partial Continuity vs. Full Continuity

A function can be continuous in one variable but not in the other. For example:

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

For any fixed value of $y$, $f(x, y)$ is a continuous function of $x$. Similarly, for any fixed value of $x$, $f(x, y)$ is continuous in $y$. However, this does not guarantee that $f(x, y)$ is continuous everywhere in the $(x, y)$ plane.

To check full continuity, we must consider both variables together. In this case, $f(x, y)$ is actually continuous everywhere because it’s a rational function with no points where the denominator is zero.

The Role of Partial Derivatives in Continuity

Partial derivatives can help us understand continuity. If all the partial derivatives of a function exist and are continuous, then the function itself is continuous. This is a powerful result, known as the differentiability implies continuity theorem.

Example: Differentiability Implies Continuity

Consider the function:

f(x, y) = x^2 + 3xy + y^2

Let’s find the partial derivatives:

f_x(x, y) = 2x + 3y, \quad f_y(x, y) = 3x + 2y

Both $f_x(x, y)$ and $f_y(x, y)$ are polynomials, and they are continuous everywhere. Therefore, $f(x, y)$ is continuous everywhere.

When Partial Derivatives Don’t Guarantee Continuity

However, the existence of partial derivatives alone doesn’t guarantee continuity. For example:

$f(x, y) = \begin{cases}$

$\frac{xy}{x^2 + y^2}$ & \text{if } (x, y) $\neq$ (0,0) \\

0 & \text{if } (x, y) = (0,0)

$\end{cases}$

We’ve already found that $f(x, y)$ is continuous at $(0,0)$. But let’s check the partial derivatives at $(0,0)$.

Partial derivative with respect to $x$ at $(0,0)$:

f_x(0,0) = $\lim_{h \to 0}$ $\frac{f(h,0) - f(0,0)}{h}$ = $\lim_{h \to 0}$ $\frac{0 - 0}{h}$ = 0

Partial derivative with respect to $y$ at $(0,0)$:

f_y(0,0) = $\lim_{k \to 0}$ $\frac{f(0,k) - f(0,0)}{k}$ = $\lim_{k \to 0}$ $\frac{0 - 0}{k}$ = 0

So, the partial derivatives exist at $(0,0)$, but they don’t tell us about continuity because this function was defined piecewise. Always check the full definition of the function!

Conclusion

In this lesson, you’ve explored the concepts of limits and continuity in multivariable calculus. You learned how to evaluate limits using different paths and polar coordinates, and you discovered the conditions for a function to be continuous in two dimensions. We also looked at examples of functions that are continuous and those that are not, and we examined the role of partial derivatives in determining continuity. With these tools, you’re well-equipped to handle the challenges of limits and continuity in higher dimensions. Keep practicing, and you’ll master these concepts in no time! 🌟

Study Notes

A limit in two variables:

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

means $f(x, y)$ approaches $L$ as $(x, y)$ approaches $(a, b)$ from all paths.

The limit must be the same along every possible path for it to exist.

Polar coordinates:

x = r $\cos$ $\theta$, \quad y = r $\sin$ $\theta$

are useful for evaluating limits. In polar form, as $r \to 0$, the behavior of the function can be analyzed more easily.

If different paths yield different limits, the limit does not exist.

Continuity conditions:

$f(a, b)$ is defined.
$\lim_{(x,y) \to (a,b)} f(x,y)$ exists.
$\lim_{(x,y) \to (a,b)} f(x,y) = f(a, b)$.

A function is continuous on its domain if it’s continuous at every point in that domain.

Piecewise functions can be continuous if the limit at the boundary point matches the defined value of the function.

Partial derivatives can help check continuity:
If partial derivatives exist and are continuous, the function is continuous.
However, the existence of partial derivatives alone does not guarantee continuity.

Differentiability implies continuity: If $f(x, y)$ is differentiable at $(a, b)$, then $f(x, y)$ is continuous at $(a, b)$.

Keep these ideas in mind, and you’ll be ready to tackle limits and continuity problems with confidence! 🚀