Problem 19 - Calculus with Concepts in Calculus

Question

Use the definition of \(\lim _{(x, y)_{\vec{R}}\left(x_{0}, y_{0}\right)} f(x, y)\) to determine whether the given limit exists for the given region \(R\). If the limit exists, find it. \(\lim _{(x, y) \vec{R}(0,0)} \frac{\sin (x-y)}{x-y} ; R\) consists of all \((x, y)\) such that \(x \neq y\)

Step-by-Step Solution

Answer

The limit exists and is 1.

Step 1: Understand the Problem Statement

We need to determine whether the limit \(\lim _{(x, y) \to (0,0)} \frac{\sin (x-y)}{x-y}\) exists when approaching (0,0) through points in the region \(R\) where \(x eq y\). If it exists, we must find the limit value.

Step 2: Consider the Path Approaching the Limit

The function \(\frac{\sin (x-y)}{x-y}\) resembles the standard limit \(\lim _{u \to 0} \frac{\sin u}{u} = 1\). As we approach \((0, 0)\), we want \((x-y)\) to approach 0 while ensuring \(x eq y\). Because \(\sin(u)/u\) approaches 1, we predict this is the outcome, given that \(x eq y\), meaning \((x-y) eq 0\), just right at the point.

Step 3: Apply Epsilon-Delta Definition

According to the epsilon-delta definition of a limit, for every \(\epsilon > 0\), there must be a \(\delta > 0\) such that if \(0 < \sqrt{x^2 + y^2} < \delta\) (and \(x eq y\)), then \(\left| \frac{\sin(x-y)}{x-y} - 1 \right| < \epsilon\). As \((x,y)\) approaches \((0,0)\), this condition holds true by the reasoning of \(\lim_{u \to 0} \frac{\sin u}{u} = 1\).

Step 4: Conclusion

Since the function behaves as expected along all paths in the region \(R\) where \(x eq y\), the limit exists and equals 1 based on analogous limit behavior in one variable calculus.

Key Concepts

epsilon-delta definition

The epsilon-delta definition is a fundamental concept in calculus used to define the limit of a function. It's particularly powerful because it provides a formal way to discuss the behavior of functions as they approach a certain point. Here's how it works in simple terms:

• For a function \(f(x, y)\) and a point \((x_0, y_0)\), we say that \(L\) is the limit of \(f(x, y)\) as \( (x, y)\) approaches \((x_0, y_0)\), if for every \(\epsilon > 0\) (no matter how small), there exists a \(\delta > 0\) such that whenever \(0 < \sqrt{(x-x_0)^2 + (y-y_0)^2} < \delta\), then \(|f(x, y) - L| < \epsilon\).
• This condition ensures that as your points \((x, y)\) get closer to \((x_0, y_0)\), the function values get arbitrarily close to the value \(L\).

This definition is crucial when examining functions of multiple variables because it helps rigorously determine whether a limit at a particular point exists.

limit using paths

To determine if a limit exists, one useful method is using paths. Paths offer a way of assessing how a function behaves as it nears a given point along different trajectories.

• In multivariable calculus, you can approach a point by moving along different lines or curves in the input space.
• If, along every possible path leading to the given point, the function approaches the same limiting value, the limit exists and is this value.
• Conversely, if different paths yield different limits, the limit does not exist.

Connecting this to our exercise, as we take paths where \(x eq y\) towards \((0,0)\), like any line \(y = x + m\), the function \(\frac{\sin(x-y)}{x-y}\) always approaches 1, suggesting that the limit is indeed 1 as long as \(x eq y\).

sine function limits

The sine function allows us to explore interesting behavior in limits, particularly with the classic limit \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \). This is a pivotal result in calculus, offering insight into trigonometric functions' local linearity.

• This specific ratio behaves very similarly to the identity function near zero, meaning as \(u\) becomes very small, \(\frac{\sin u}{u}\) becomes very close to 1.
• This concept extends naturally to multivariable calculus when dealing with functions like \(\frac{\sin(x-y)}{x-y}\), as explored in the exercise.
• When \((x-y)\) approaches zero (but not exactly zero due to the domain restriction \(x eq y\)), this ratio also tends to behave like our single-variable result, approaching the limit of 1.

Understanding how this foundational limit behaves is key to handling more complex multivariable limits.

multivariable calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of more than one variable. This branch of mathematics is essential for understanding changes in systems described by several variables.

• It studies functions \(f(x, y, z, \ldots)\). These are often explored in terms of limits, derivatives, and integrals just like single-variable functions, but with additional complexities.
• Calculating limits in multivariable calculus introduces new challenges, such as ensuring that the limit is the same regardless of the path taken to approach the point.
• Exercises like our given problem demonstrate how concepts like the epsilon-delta definition and path independence are operationalized in this context. Here, understanding that the limit exists requires verifying it through various paths where \(x eq y\).

Unraveling multivariable calculus concepts helps in modeling real-world phenomena involving several interdependent quantities.