Problem 72 - 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. $$\lim _{(x, y) \rightarrow(0,0)} \sqrt{9-x^{2}-y^{2}}$$

Step-by-Step Solution

Answer

The limit is 3.

Step 1: Identify the Expression

We need to evaluate the limit \( \lim _{(x, y) \rightarrow(0,0)} \sqrt{9-x^{2}-y^{2}} \).

Step 2: Analyze the Domain

The expression \( \sqrt{9-x^2-y^2} \) is only defined where \( 9 - x^2 - y^2 \geq 0 \). This is equivalent to \( x^2 + y^2 \leq 9 \), which is the equation of a circle with radius 3 centered at the origin. This ensures the expression is defined in a certain region around \((0,0)\).

Step 3: Plug in the Limit Point Values

Substitute \(x = 0\) and \(y = 0\) into the expression: \(\sqrt{9 - 0^{2} - 0^{2}} = \sqrt{9} = 3.\)

Step 4: Consider Path-Independent Approach

The given function appears to be path-independent because it involves simple polynomials under the square root. The form of \(9-x^2-y^2\) suggests the limit is evaluated consistently toward \((0,0)\) from any path inside the circle \(x^2 + y^2 < 9\).

Step 5: Conclude the Limit's Existence

The limit \( \lim _{(x, y) \rightarrow(0,0)} \sqrt{9-x^{2}-y^{2}} \) is 3 because it is consistent despite the chosen path towards \((0,0)\) within its domain. The function resolves to a single constant that does not depend on \(x\) or \(y\) when \(x,y\rightarrow 0,0\).

Key Concepts

Limit Evaluation

In multivariable calculus, evaluating limits involves finding the behavior of a function as the input variables approach a particular point. In our example, we are tasked with finding the limit of the function \(\lim _{(x, y) \rightarrow(0,0)} \sqrt{9-x^{2}-y^{2}}\). The goal is to check if the function approaches a specific value as \(x\) and \(y\) get closer to \(0\). If this value is the same regardless of the path taken to approach \((0,0)\), then the limit exists.

Understanding limit evaluation:

• Identify the function to be evaluated as the variables approach a certain point.
• Consider if the same result is achieved from all possible paths approaching that point.
• If the result varies based on the path, the limit does not exist.

In our exercise, substituting \(x = 0\) and \(y = 0\) directly, we find the expression simplifies neatly to \(3\), indicating that the limit is \(3\) as \((x, y)\) approaches \((0, 0)\).

Domain Analysis

Analyzing the domain of a function is crucial to understanding where the function is defined and meaningful. In our exercise, the function is \(\sqrt{9-x^2-y^2}\). This expression is only valid if what's under the square root is non-negative, meaning:

• The inequality \(9 - x^2 - y^2 \geq 0\) must be true.
• This represents a circular region centered at the origin with radius 3.

Thus, the domain of our function is all points \((x, y)\) that lie within or on the boundary of this circle. Understanding this domain is key because we need to ensure our limit point \((0, 0)\) lies within this valid region. Remember, the function isn't valid beyond this circular boundary, so approaching from there wouldn't make sense.

Path-Independence

Path-independence in limits is an important aspect to consider in multivariable functions. A function is path-independent regarding its limits if the approach to the point yields the same result regardless of the path taken.

For our function \(\lim _{(x, y) \rightarrow(0,0)} \sqrt{9-x^{2}-y^{2}}\), path-independence suggests that:

• The result is consistent along all valid pathways within the circle \(x^2 + y^2 < 9\).
• This consistency is possible because the expression under the square root simplifies path-independently.

This means we don't have to check every possible path, as the form of our function guarantees the limit will be the same when approaching \((0, 0)\) from any direction inside the domain.

Function Continuity

Function continuity means that a function does not have any abrupt changes or jumps in its value. For a function to be continuous at a point, the limit of the function as it approaches the point from any direction must be equal to the function's value at that point.

In our exercise, the function \(\sqrt{9-x^{2}-y^{2}}\) is continuous at \((0, 0)\). All paths towards this point within the circle lead to a limit of \(3\), which matches the function's value at \((0, 0)\), namely, \(3\).
As part of checking continuity:

• Ensure the function's limit equals the function's value at the point of interest.
• Verify that this holds from all paths in its domain reaching the point.

For our exercise, the function resolves without any sudden leaps or gaps at the center \((0, 0)\), confirming its continuity at this location.