Question

In Exercises \(25-28,\) discuss the continuity of the function and evaluate the limit of \(f(x, y)\) (if it exists) as \((x, y) \rightarrow(0,0)\). \(f(x, y)=e^{x y}\)

Short answer

The function \(f(x, y) = e^{xy}\) is continuous at \((0,0)\) and the limit as \((x, y)\) approaches \((0, 0)\) is \(1\).

Step-by-step solution

Determining the Continuity

The continuity of a function at a point requires the function to be defined at that point and the limit to exist at that point. The function \(f(x, y) = e^{xy}\) is defined for all real values of \(x\) and \(y\), and there are no discontinuities.

Finding the Limit

The limit of a function as \((x, y)\) approaches a point is the value the function approaches as \((x, y)\) approaches that point. To find \(\lim_{{(x, y) \to (0,0)}} f(x,y)\), we substitute \(0\) for \(x\) and \(y\): \[ \lim_{{(x, y) \to (0,0)}} e^{xy} = e^{(0)(0)} = e^{0} = 1.\]

Key concepts

Continuity of Functions

Understanding the continuity of functions in multivariable calculus is a cornerstone for analyzing complex mathematical landscapes. A function like f(x, y) = e^xy is considered continuous if, at every point in its domain, the following two criteria are satisfied: the function must be defined, and the limit must exist and equal the function's value at that point.

For our function e^xy, it is defined for all real numbers of x and y. This means that no matter which values we substitute into x and y, we will always get an output. In order to confirm its continuity at the point (0,0), we evaluate the limit as (x, y) approaches this specific point. If the limit exists and equals f(0,0), which in this case is e⁰ = 1, we can conclude the function is continuous at (0,0). As we can see, no point in the function's domain produces a disruption or gap, and thus, the function is continuous everywhere.

Limit Evaluation

When evaluating the limit of a multivariable function, one is interested in finding out what value the function approaches as the input variables approach certain values. Limit evaluation in multivariable calculus can be trickier than in single-variable calculus because the path taken to reach the point of interest can affect the limit value.

However, for the function f(x, y) = e^xy, we are in a simpler situation as the function is well-behaved at the origin (0,0). To evaluate the limit as (x, y) approaches (0,0), we observe what happens when we plug in x=0 and y=0. The result is e⁽⁰⁾⁽⁰⁾ = e⁰ = 1, which is a clear indication of the function's behavior at that point. Given this, we can confidently say that the limit of f(x, y) as (x, y) approaches (0,0) is indeed 1.

Multivariable Calculus

Multivariable calculus extends the principles of single-variable calculus to functions of several variables. Concepts such as limits, continuity, derivatives, and integrals are expanded upon to accommodate the complexity of paths in a multidimensional context. In the study of functions such as f(x, y) = e^xy, it is essential to consider each variable's interaction and combined effect on the function's value.

From representing surfaces in three-dimensional space to calculating rates of change that depend on multiple factors, multivariable calculus offers the mathematical framework necessary for modeling and solving real-world problems involving multiple variables. With a multivariable function, the route you take to approach a certain point can lead to different limit values, yet if a function is continuous, like our exponential function, the limit is robust and remains constant irrespective of the path taken, which greatly simplifies the analysis and problem-solving processes.