Problem 42 - Calculus Early Transcendentals

Question

At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$f(x, y)=e^{x^{2}+y^{2}}$$

Step-by-Step Solution

Answer

Answer: The function \(f(x,y) = e^{x^2+y^2}\) is continuous at every point (x, y) in the 2-dimensional real plane \(\mathbb{R}^2\).

Step 1: Recall the definition of continuity

A function is continuous at a point (a, b) if the following conditions are met: 1. The function is defined at (a, b). 2. The limit of the function as (x, y) approaches (a, b) exists. 3. The limit of the function as (x, y) approaches (a, b) equals the value of the function at (a, b). Mathematically, this can be written as: $$\lim_{(x, y) \to (a, b)} f(x, y) = f(a, b)$$ We need to check whether the given function \(f(x, y) = e^{x^2+y^2}\) satisfies these conditions for every point (x, y) in \(\mathbb{R}^2\).

Step 2: Check if the function is defined at all points in \(\mathbb{R}^2\)

The function we are given, \(f(x, y) = e^{x^2+y^2}\), is an exponential function. Exponential functions are always defined for any real input, and in this case, the input consists of x and y. Therefore, the function is indeed defined at all points in \(\mathbb{R}^2\).

Step 3: Check if the limit of the function exists

In order to check if the limit exists, we can look at the function terms \(x^2+y^2\). Since both of the terms are squared, their values will always be non-negative. Therefore, the input to the exponential function (x^2+y^2) will range from \(0\) to positive infinity. The exponential function is continuous and defined for all real input values, so as long as the input \((x^2 + y^2)\) is continuous, the output \(e^{x^2+y^2}\) will also be continuous. The limit of the function as (x, y) approaches (a, b) is guaranteed to exist, since the exponential function behaves nicely for all real input values.

Step 4: Show that the limit equals the function value

From Step 3, we know that the limit of the function as (x, y) approaches (a, b) exists. We have: $$\lim_{(x, y) \to (a, b)} e^{x^2+y^2} = e^{a^2+b^2}$$ The right-hand side of the equation above is exactly the value of the function at the point (a, b), which is \(f(a, b)\). Therefore, the limit of the function as (x, y) approaches (a, b) equals the value of the function at (a, b).

Step 5: Conclude the function continuity

Since the function \(f(x, y) = e^{x^2+y^2}\) is: 1. Defined at all points in \(\mathbb{R}^2\) 2. The limit of the function as (x, y) approaches (a, b) exists 3. The limit of the function as (x, y) approaches (a, b) equals the value of the function at (a, b) we can conclude that the function is continuous at every point (x, y) in the 2-dimensional real plane \(\mathbb{R}^2\).