Problem 71
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
Verified Answer
The limit is \( e^{-32} \).
1Step 1: Understand the Function and the Problem
The function given is \( e^{-x^2 - y^2} \). We need to evaluate the limit of this function as \((x, y)\) approaches \((4, 4)\). This means we observe the behavior of the function's value as \(x\) and \(y\) get very close to 4.
2Step 2: Substitute the Values into the Function
Substitute \(x = 4\) and \(y = 4\) into the function. Doing this gives us the expression \( e^{-(4)^2 - (4)^2} \).
3Step 3: Simplify the Expression
Calculate \(4^2\) to get 16. Then substitute back into the function to get \( e^{-16 - 16} \), which simplifies to \( e^{-32} \).
4Step 4: Determine the Limit Value
The expression \( e^{-32} \) is the limit. Since \( e^{-32} \) results in a very small positive number (close to zero), the function approaches this value as \((x, y)\) approaches \((4, 4)\). There are no undefined operations or discontinuities in the expression, so the limit exists.
Key Concepts
Exponential FunctionLimit EvaluationContinuity of Functions
Exponential Function
The exponential function, denoted as \( e^{x} \), is a mathematical function of the form \( f(x) = e^{x} \). Here, \( e \) represents Euler's number, approximately equal to 2.71828. It is a unique function because it has a constant rate of growth proportional to its current value.
Exponential functions often appear in various scientific and mathematical contexts due to their characteristic of representing life-like processes, such as population growth or radioactive decay. In our exercise, the function is \( e^{-x^2 - y^2} \). This particular form is a two-variable function where both \( x \) and \( y \) impact the output simultaneously.
The negative sign in \( -x^2-y^2 \) ensures the result is a descending exponential value since squaring \( x \) and \( y \) will always yield positive terms. Therefore, \( e^{-x^2-y^2} \) represents a surface that decreases rapidly as \( x^2 + y^2 \) grows, making it handy for modeling diffusion or heat distribution scenarios in two-dimensional spaces.
Exponential functions often appear in various scientific and mathematical contexts due to their characteristic of representing life-like processes, such as population growth or radioactive decay. In our exercise, the function is \( e^{-x^2 - y^2} \). This particular form is a two-variable function where both \( x \) and \( y \) impact the output simultaneously.
The negative sign in \( -x^2-y^2 \) ensures the result is a descending exponential value since squaring \( x \) and \( y \) will always yield positive terms. Therefore, \( e^{-x^2-y^2} \) represents a surface that decreases rapidly as \( x^2 + y^2 \) grows, making it handy for modeling diffusion or heat distribution scenarios in two-dimensional spaces.
Limit Evaluation
Limit evaluation is a fundamental concept in calculus that describes the value a function approaches as the inputs (such as \( x \) and \( y \)) approach a certain point. In dealing with multivariable functions, such evaluations can initially seem challenging, but they follow systematic methods.
Let's consider the limit in our exercise: \( \lim_{(x, y) \rightarrow (4,4)} e^{-x^{2} - y^{2}} \). Evaluating this limit requires substituting \( x = 4 \) and \( y = 4 \) into the function then simplifying:
This result, \( e^{-32} \), is very close to zero but remains positive as \( e \) is always a positive number, even for large negative exponents. Therefore, the limit is a defined, existing small number.
Let's consider the limit in our exercise: \( \lim_{(x, y) \rightarrow (4,4)} e^{-x^{2} - y^{2}} \). Evaluating this limit requires substituting \( x = 4 \) and \( y = 4 \) into the function then simplifying:
- First, calculate \( 4^2 \) for both \( x \) and \( y \), yielding 16 for each.
- Then, combine them to get \( e^{-16 - 16} \) which simplifies to \( e^{-32} \).
This result, \( e^{-32} \), is very close to zero but remains positive as \( e \) is always a positive number, even for large negative exponents. Therefore, the limit is a defined, existing small number.
Continuity of Functions
Continuity is a key aspect when examining limits because it determines if a function behaves smoothly near a certain point. A function is continuous at a point if the following conditions are met:
In the given exercise, as \((x, y)\) approaches \((4, 4)\), \( e^{-x^2 - y^2} \) smoothly approaches a single limit value \( e^{-32} \). There are no discontinuities or undefined operations, meaning that the function behaves predictably and continuously around \((4, 4)\). This property simplifies limit evaluations, allowing you to directly substitute values in the function, reinforcing the concept's importance in practical applications.
- The function is defined at the point.
- The limit exists at that point.
- The limit equals the function's value at that point.
In the given exercise, as \((x, y)\) approaches \((4, 4)\), \( e^{-x^2 - y^2} \) smoothly approaches a single limit value \( e^{-32} \). There are no discontinuities or undefined operations, meaning that the function behaves predictably and continuously around \((4, 4)\). This property simplifies limit evaluations, allowing you to directly substitute values in the function, reinforcing the concept's importance in practical applications.
Other exercises in this chapter
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