Problem 14 - Calculus with Concepts in Calculus

Question

Evaluate the limit. $$ \lim _{(x, y) \rightarrow(0,0)} \frac{x y}{\sqrt{x^{2}+y^{2}}} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understanding the Problem

We need to find the limit of the function \( f(x, y) = \frac{xy}{\sqrt{x^2 + y^2}} \) as \((x, y)\) approaches \((0, 0)\). This involves analyzing whether the limit exists and, if it does, determining its value.

Step 2: Convert to Polar Coordinates

To evaluate the limit, convert the Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\), where \(x = r\cos\theta\) and \(y = r\sin\theta\). The distance \( \sqrt{x^2 + y^2} = r \). Substitute in the function: \( \frac{(r\cos\theta)(r\sin\theta)}{r} = r\cos\theta\sin\theta \).

Step 3: Simplify the Expression

The expression simplifies to \( r\cos\theta\sin\theta \). We notice that the variable that dominates the behavior of this expression as \(r\) approaches zero is \(r\), since \(\cos\theta\sin\theta\) is bounded.

Step 4: Evaluate the Limit as r Approaches Zero

As \(r\) approaches zero, \(r\cos\theta\sin\theta\) tends to zero because \(r\) multiplies a bounded function \(\cos\theta\sin\theta\). Therefore, the limit of the expression is 0 regardless of the direction (value of \(\theta\)) taken to approach the origin. Thus, \(\lim _{r \to 0} r \cdot \cos\theta \cdot \sin\theta = 0\).

Step 5: Conclude the Calculation

Since we have shown that the limit approaches the same value of 0 in all directions, the limit of the function \( \lim _{(x, y) \rightarrow(0,0)} \frac{xy}{\sqrt{x^2 + y^2}} \) is indeed 0.

Key Concepts

Limits in Multiple Dimensions

When we talk about limits in multiple dimensions, we're exploring how a function behaves as the input variables approach a certain point. Here, we are interested in understanding the behavior of functions with two or more variables, like our function \( f(x, y) = \frac{xy}{\sqrt{x^2 + y^2}} \), when both \(x\) and \(y\) move towards zero.

Unlike single-variable limits, where we check values as a number approaches a point from left and right, multivariable limits involve checking from infinitely many paths toward the destination point. To confirm that a limit exists, it must approach the same value, regardless of the path taken.

For example, if different paths lead to different limit values, then the limit does not exist. In our multivariable world, proving the limit exists means demonstrating consistency across all these tiny imaginary paths.

Polar Coordinates

Polar coordinates are an alternative to Cartesian coordinates that can simplify working with functions in multiple dimensions, especially when dealing with circular or radial symmetry.

In our exercise, we convert \((x, y)\) into polar coordinates \((r, \theta)\), where:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)
• The distance \( \sqrt{x^2 + y^2} = r \)

This transformation allows us to express the function as \( \frac{(r\cos\theta)(r\sin\theta)}{r} = r\cos\theta\sin\theta \).

Using polar coordinates can be particularly helpful, as it often reduces the complexity of a two-variable function into a simpler form by focusing on the radius \(r\) and the angle \(\theta\). This approach makes it easier to evaluate limits, especially when approaching the origin or points with radial symmetry.

Bounded Functions

A bounded function is a function with values that do not exceed certain limits no matter the input. Understanding this is crucial when evaluating limits.

In our problem, the component \( \cos\theta\sin\theta \) is bounded. This means for all values of \(\theta\), the values of \(\cos\theta\sin\theta\) are limited within a specific range. Specifically, since \(-1 \leq \cos\theta, \sin\theta \leq 1\), the product also remains between \(-1/2\) and \(1/2\).

When a term \(r\) tends to zero, multiplying it by a bounded function (like \(\cos\theta\sin\theta\)) results in the entire expression tending towards zero. This property helps us in step 4 of the solution. Hence, any part of the function that is multiplying \(r\) doesn't influence the limit because \(r\) dominates it by approaching zero itself, proving the overall limit exists and equals zero.