Problem 71 - Calculus Volume 3

Question

For the following exercises, evaluate the limits at the indicated values of \(x\) and \(y\). If the limit does not exist, state this and explain why the limit does not exist. $$\lim _{(x, y) \rightarrow(4,4)} e^{-x^{2}-y^{2}}$$

Step-by-Step Solution

Answer

The limit is \( e^{-32} \).

Step 1: Understanding the Expression

We need to evaluate the limit of the function \( e^{-x^2 - y^2} \) as \( (x, y) \) approaches \( (4, 4) \). The expression inside the exponential is \( -x^2 - y^2 \).

Step 2: Substitute Values into the Expression

Substitute \( x = 4 \) and \( y = 4 \) into the expression \( -x^2 - y^2 \). This yields \( -(4)^2 - (4)^2 = -16 - 16 = -32 \).

Step 3: Evaluate the Exponential Function

Substitute \( -32 \) into the exponential function to get \( e^{-32} \).

Step 4: Determine Limit Conclusion

Since \( e^{-32} \) is a defined real number, the limit of \( e^{-x^2 - y^2} \) exists and is equal to \( e^{-32} \).

Key Concepts

Limits in Multivariable Calculus

Limits in multivariable calculus involve functions of more than one variable, typically noted as functions of the form \( f(x, y) \). The main goal is to determine the behavior of \( f(x, y) \) as the input values \( (x, y) \) approach a certain point, such as \( (a, b) \). This is similar to finding limits in single-variable calculus, but the primary challenge is that there are infinitely many paths approaching \( (a, b) \).

To evaluate these limits, you should:

• Understand the behavior of the function along different paths to the point of interest.
• Check if the function approaches the same limit along all those paths.
• Conclude that if the behavior is consistent across all paths, the limit exists. Otherwise, it does not exist if it varies.

By determining a limit in multivariable calculus, you are affirming the stability of the function around the point and ensuring that variables behave consistently as they change.

Exponential Functions

Exponential functions like \( e^{f(x, y)} \) play a key role in both single- and multivariable calculus, characterized by the constant base \( e \) raised to the power of a given expression \( f(x, y) \). These functions are continuous everywhere, which is a vital property when dealing with limits.

When considering multivariable exponential functions:

• The exponent can be any real-valued function of the variables \( x \) and \( y \).
• If \( f(x, y) \) has a limit as \( (x, y) \to (a, b) \), then \( e^{f(x, y)} \) also has a limit as the same point.
• This is because the exponential function is continuous, meaning it smoothly follows changes in its input without breaking or jumping.

Using exponential functions in calculus allows for modeling growth and decay, among other phenomena, where understanding limits helps predict how these processes behave near specific points.

Evaluating Limits

When evaluating limits, especially in the context of multivariable functions, the process involves substituting values into the function to see how it behaves near the target point. For the limit \( \lim_{(x, y) \to (4, 4)} e^{-x^2-y^2} \), let's break down the evaluation method.

First, begin by substituting both \( x = 4 \) and \( y = 4 \) directly into the expression inside the exponential. Evaluate \( -x^2-y^2 \) by computation:

• Calculate \( 4^2 + 4^2 \), which results in \( 16 + 16 = 32 \).
• Thus, \( -(4^2+4^2) = -32 \).

Now, insert this value back into the exponential function:

• Evaluate \( e^{-32} \). Since this is a finite number and part of the continuous range of the exponential function, the limit \( \lim_{(x, y) \to (4, 4)} e^{-x^2-y^2} \) is \( e^{-32} \).

The existence of such a limit is determined by the continuity and well-defined nature of the exponential function, confirming a stable behavior near the point of interest.