Let f(x, y) be differentiable at the point (x0, y0). Prove that f(x, y) is continuous at (x0, y0). Hint: Consider the function ε(∆x, ∆y) = f(x0 + ∆x, y0 + ∆y) − f(x0, y0) − ∆xfx(x0, y0) − ∆yfy(x0, y0) p (∆x) 2 + (∆y) 2

To prove that \( f(x, y) \) is continuous at \( (x_0, y_0) \), we need to show that:

\[
\lim_{(x, y) \to (x_0, y_0)} f(x, y) = f(x_0, y_0)
\]

Given that \( f(x, y) \) is differentiable at \( (x_0, y_0) \), we know the following definition holds:

\[
f(x_0 + \Delta x, y_0 + \Delta y) = f(x_0, y0) + \Delta x f_x(x_0, y_0) + \Delta y f_y(x_0, y_0) + \epsilon(\Delta x, \Delta y)
\]

Where \( \epsilon(\Delta x, \Delta y) \) is a function that satisfies:

\[
\lim_{(\Delta x, \Delta y) \to (0, 0)} \frac{\epsilon(\Delta x, \Delta y)}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} = 0
\]

This property implies that the function \( \epsilon(\Delta x, \Delta y) \) tends to zero faster than \( \sqrt{(\Delta x)^2 + (\Delta y)^2} \) as \( (\Delta x, \Delta y) \to (0, 0) \).

### Proof of Continuity:

Let's rewrite the expression for \( f(x_0 + \Delta x, y_0 + \Delta y) \) using \( h = \sqrt{(\Delta x)^2 + (\Delta y)^2} \) and observe the following:

\[
f(x_0 + \Delta x, y_0 + \Delta y) = f(x_0, y_0) + \Delta x f_x(x_0, y_0) + \Delta y f_y(x_0, y_0) + \epsilon(\Delta x, \Delta y)
\]

Now, express the above equation as:

\[
f(x_0 + \Delta x, y_0 + \Delta y) - f(x_0, y_0) = \Delta x f_x(x_0, y_0) + \Delta y f_y(x_0, y_0) + \epsilon(\Delta x, \Delta y)
\]

Taking the limit as \( (\Delta x, \Delta y) \to (0, 0) \):

\[
\lim_{(\Delta x, \Delta y) \to (0, 0)} \left[ f(x_0 + \Delta x, y_0 + \Delta y) - f(x_0, y_0) \right] = \lim_{(\Delta x, \Delta y) \to (0, 0)} \left[ \Delta x f_x(x_0, y_0) + \Delta y f_y(x_0, y_0) + \epsilon(\Delta x, \Delta y) \right]
\]

Since \( \Delta x \) and \( \Delta y \) both approach 0, the linear terms \( \Delta x f_x(x_0, y_0) \) and \( \Delta y f_y(x_0, y_0) \) will also approach 0. Moreover, by the property of \( \epsilon(\Delta x, \Delta y) \):

\[
\lim_{(\Delta x, \Delta y) \to (0, 0)} \epsilon(\Delta x, \Delta y) = 0
\]

Thus:

\[
\lim_{(\Delta x, \Delta y) \to (0, 0)} \left[ f(x_0 + \Delta x, y_0 + \Delta y) - f(x_0, y_0) \right] = 0
\]

This implies:

\[
\lim_{(x, y) \to (x_0, y_0)} f(x, y) = f(x_0, y_0)
\]

Hence, \( f(x, y) \) is continuous at \( (x_0, y_0) \).

Would you like further details or have any questions about this proof?

Here are some related questions:
1. What does it mean for a function to be differentiable at a point?
2. How does differentiability imply continuity in general?
3. Can a function be continuous but not differentiable? Provide an example.
4. What are partial derivatives, and how are they used in multivariable calculus?
5. What is the significance of the function \( \epsilon(\Delta x, \Delta y) \) in the context of differentiability?

**Tip:** When proving the continuity of a function at a point, consider the limit definition of continuity and explore the connection to differentiability, as it often simplifies the proof.

Explore how to prove continuity from differentiability at a given point in multivariable calculus. This detailed mathematical proof involves the limit definition of continuity and the connection to differentiability. Suitable for advanced undergraduate students and learners of multivariable calculus.

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