Problem 37 - Calculus Early Transcendentals

Question

\(29-38\) Determine the set of points at which the function is continuous. $$f(x, y)=\left\\{\begin{array}{ll}{\frac{x^{2} y^{3}}{2 x^{2}+y^{2}}} & {\text { if }(x, y) \neq(0,0)} \\ {1} & {\text { if }(x, y)=(0,0)}\end{array}\right.$$

Step-by-Step Solution

Answer

The function is continuous everywhere except at the origin (0, 0).

Step 1: Analyze Continuity at Non-origin Points

For points \((x, y) eq (0, 0)\), analyze the continuity of the function \(f(x, y) = \frac{x^2 y^3}{2x^2 + y^2}\). This expression is a rational function and is continuous wherever the denominator is non-zero. Since \(2x^2 + y^2 > 0\) for \((x, y) eq (0, 0)\), the function is continuous at all such points.

Step 2: Examine the Origin for Continuity

At the origin \((x, y) = (0, 0)\), the function is defined differently: \(f(0, 0) = 1\). To check continuity here, evaluate the limit \(\lim_{(x, y) \to (0, 0)} f(x, y)\). If this limit equals \(f(0, 0) = 1\), then \(f\) is continuous at the origin.

Step 3: Calculate Limit Along x-axis

Evaluate the limit along the x-axis, where \(y = 0\):\[\lim_{x \to 0} f(x, 0) = \lim_{x \to 0} \frac{x^2 \cdot 0^3}{2x^2 + 0^2} = \lim_{x \to 0} 0 = 0.\]Thus, the limit along the x-axis is 0.

Step 4: Calculate Limit Along y-axis

Evaluate the limit along the y-axis, where \(x = 0\):\[\lim_{y \to 0} f(0, y) = \lim_{y \to 0} \frac{0^2 y^3}{2\cdot 0^2 + y^2} = \lim_{y \to 0} 0 = 0.\]Thus, the limit along the y-axis is 0.

Step 5: Calculate Limit Along y = x

Evaluate the limit along the line \(y = x\):\[\lim_{x \to 0} f(x, x) = \lim_{x \to 0} \frac{x^2 x^3}{2x^2 + x^2} = \lim_{x \to 0} \frac{x^5}{3x^2} = \lim_{x \to 0} \frac{x^3}{3} = 0.\]Thus, the limit along the line \(y = x\) is 0.

Step 6: Conclude Continuity Properties

Since the limit of \(f(x, y)\) as \((x, y) \to (0, 0)\) is 0 along multiple paths and these do not equal \(f(0, 0) = 1\), the limit is different from the function value at the origin. Hence, \(f\) is not continuous at \((0, 0)\). The function is continuous for all \((x, y) eq (0, 0)\).

Key Concepts

Limits in Multivariable Calculus

When studying limits in multivariable calculus, it's essential to understand how values approach a particular point in a space with more than one dimension, like in 2D or 3D spaces. In this multivariable context, we explore how a function behaves as both input variables approach certain values simultaneously. For example, as both \(x\) and \(y\) move closer to zero, we analyze the limit of the function \(f(x, y) = \frac{x^2 y^3}{2x^2 + y^2}\). The expression is evaluated by considering various paths toward the point, such as along the x-axis, y-axis, or along the line \(y = x\).

• By checking limits along different paths, we ensure that the function approaches the same value regardless of the direction of approach.


• If these limits differ, it indicates the function does not have a limit at that point, and thus cannot be continuous there.

Determining the behavior of multivariable functions often involves a graphical representation and analytical computation, making it a crucial stepping stone in understanding continuity.

Continuity in Calculus

Continuity is a fundamental concept in calculus. A function is continuous at a point if its limit exists as it approaches that point, and this limit matches the function's value at the point. For a multivariable function, continuity adds complexity because both the limit and function value should be consistent regardless of the path taken by the input variables.
In our exercise, we defined the function \(f(x,y)\) differently at the origin, \(f(0,0) = 1\), than at other points. To test continuity at \((0,0)\), we checked the above-mentioned limit :

• Calculate limits along different approaches to \((0,0)\), such as x-axis, y-axis, and the line \(y = x\).


• Compare these path-specific limits to the defined function value at the origin.

Here, although the limits along chosen paths approach 0, the function's value is 1 at the origin, signifying discontinuity. This analysis highlights just how dynamic and path-dependent continuity assessments have to be in multivariable calculus.

Piecewise Functions

A piecewise function is defined by different expressions for different parts of its domain. This makes it flexible, as it can handle specific cases separately within the same function. In our example, the function \(f(x,y)\) is expressed as a piecewise function:

• One expression for points \((x, y) eq (0,0)\).


• Another single value (1) when \((x, y) = (0,0)\).

This helps tailor the function's behavior at specific, potentially complex points such as the origin, where typical function expressions might not apply or lead to undefined outcomes.
When analyzing piecewise functions, especially for continuity, attention is drawn towards these transition points or regions where the function definition changes. Continuity checks across these areas include confirming that both the function's expression and the piecewise assignment align, ensuring a smooth transition, like a seamless path over a bridge connecting two islands. In multivariable calculus, this requires rigorous approach-path analysis for any non-density-causing points in the domain.