Problem 15 - The Calculus with Analytic Geometry

Question

In Exercises 13 through 16, prove that \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\) exists. \(f(x, y)= \begin{cases}(x+y) \sin \frac{1}{x} \sin \frac{1}{y} & \text { if } x \neq 0 \text { and } y \neq 0 \\ 0 & \text { if either } x=0 \text { or } y=0\end{cases}\)

Step-by-Step Solution

Answer

The limit exists and is 0.

Step 1: Identify the Function Behavior

Examine the given function behavior. For points where either x or y is 0, the function value is directly given as 0, i.e., \(f(x, y)=0\) if \(x=0\) or \(y=0\).

Step 2: Evaluate along the x-axis and y-axis

Along the x-axis (\(y=0\)) and the y-axis (\(x=0\)), \(f(x, y)=0\). Thus \(\lim _{(x, y) \rightarrow (0, 0)} f(x, y) = 0\) along these axis.

Step 3: Simplify Function for x,y ≠ 0

For \(x eq 0\) and \(y eq 0\): \(f(x, y) = (x+y) \sin \frac{1}{x} \sin \frac{1}{y}\). Here, examine the possible bounds for the function.

Step 4: Bounding the Sine Function

Noting that \(|\sin \frac{1}{x}| \leq 1\) and \(|\sin \frac{1}{y}| \leq 1\), we can bound: \(|f(x, y)| = |x+y| |\sin \frac{1}{x}| |\sin \frac{1}{y}| \leq |x+y|\).

Step 5: Limit of the Bound Function

As \((x,y) \rightarrow (0,0)\), the expression \(|x+y| \rightarrow 0\). Therefore, \(|f(x, y)| \leq |x+y| \rightarrow 0\).

Step 6: Conclusion

Since for all paths leading to \((0,0)\), \(|f(x,y)|\) approaches \(0\), we conclude that \(\lim _{(x, y) \rightarrow(0,0)} f(x, y) = 0\) and exists.

Key Concepts

epsilon-delta definition

To understand the multivariable limit, we need to first grasp the \(\epsilon-\delta\) definition. This definition provides a rigorous way to define limits. For a function \(f(x,y)\) to have a limit \(L\) at a point \((a, b)\), it must satisfy the following condition: For any positive number \(\epsilon\), however small, there exists a corresponding positive number \(\delta\) such that for all \((x, y)\) within the distance \(\delta\) from \((a, b)\), \(f(x, y)\) is within the distance \(\epsilon\) from \(L\).
Mathematically, this is expressed as:
For every \(\epsilon > 0\), there exists \(\delta > 0\) such that if \(0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta\), then \(|f(x, y) - L| < \epsilon\).
This definition ensures that by making \((x, y)\) sufficiently close to \((a, b)\), the function values can be made arbitrarily close to \(L\).

bounding functions

Bounding functions help simplify complicated functions by providing an envelope within which the function's values lie. This is particularly useful when dealing with trigonometric functions like sine and cosine.
For the given function \(f(x, y) = (x+y) \, \sin \frac{1}{x} \sin \frac{1}{y}\), we notice that \(|\sin \frac{1}{x}|\) and \(|\sin \frac{1}{y}|\) are bounded by 1 since the sine function oscillates between -1 and 1.
Therefore, we can simplify things by noting that:
\(|f(x, y)| = |x+y| \, |\sin \frac{1}{x}| \, |\sin \frac{1}{y}| \, \leq |x+y|\).
By bounding the function this way, we can more easily determine the behavior of \(f(x, y)\) as \((x, y)\) approaches \((0, 0)\).

sine function behavior

Understanding the sine function is crucial to analyzing our given function. The sine function, \(\sin \theta\), oscillates between -1 and 1 for all real numbers \(\theta\). This implies that:
\(-1 \leq \sin \frac{1}{x} \leq 1\) and \(-1 \leq \sin \frac{1}{y} \leq 1\).
This periodic and bounded nature of the sine function allows us to place limits on the product of sines involved in our function.
When dealing with multivariable limits, particularly those involving trigonometric functions, this bounded behavior is beneficial. It helps in simplifying complex expressions and determining possible bounds, leading to a clearer understanding of the limit.

path independence in limits

Path independence is an essential concept in multivariable calculus. It ensures that the limit of a function as its variables approach a point is the same regardless of the path taken. For the limit \(\lim _{(x, y) \rightarrow (0, 0)} f(x, y)\) to exist, it must be path-independent.
We can test path independence by examining different paths, such as along the x-axis (\(y = 0\)), y-axis (\(x = 0\)), and linear paths (like \(x = y\)).
In the exercise, we examined:

• Along the x-axis: \(f(x, 0) = 0\)
• Along the y-axis: \(f(0, y) = 0\)

Both resulted in the function value going to 0 as \((x, y)\) approaches \((0, 0)\).
By bounding the function and noting the behavior of the sine function, we finally concluded that \(\lim _{(x, y) \rightarrow (0, 0)} |f(x, y)| \leq |x + y| \rightarrow 0\).
This path-independence assures us the limit exists and is 0.