Problem 29 - Calculus with Concepts in Calculus

Question

Let $$ f(x, y)=\left\\{\begin{array}{ll} \frac{x y^{2}}{x^{4}+y^{4}} & \text { for }(x, y) \neq(0,0) \\ 0 & \text { for }(x, y)=(0,0) \end{array}\right. $$ a. Show that \(f\) is continuous in each variable separately at \((0,0)\), that is, \(f(x, 0)\) is a continuous function of \(x\) at 0, and \(f(0, y)\) is a continuous function of \(y\) at \(0 .\) b. Show that \(f\) is not continuous at \((0,0)\).

Step-by-Step Solution

Answer

a. Yes, continuous separately in \(x\) and \(y\). b. Not continuous at \((0,0)\).

Step 1: Setting up Calculation Definitions

First, understand the function provided. It has two parts:1. For points \((x,y)\) not equal to \((0,0)\), the function is described by the equation \( f(x,y) = \frac{x y^2}{x^4 + y^4} \).2. At the point \((0,0)\), the function is defined as \( f(0,0) = 0 \).

Step 2: Part A: Considering Continuity in Terms of Each Variable Separately

For continuity in terms of \(x\) and \(y\) separately, evaluate the function as one of the variables approaches zero while the other is held constant. This reveals if it is continuous at the origin \((0,0)\).

Step 3: Evaluating \(f(x,0)\) for Continuity in \(x\) at Zero

Take the function for \(y = 0\): \[ f(x, 0) = \frac{x \cdot 0^2}{x^4 + 0^4} = 0 \].Now, as \(x\) approaches \(0\), we have:\[ \lim_{{x \to 0}} f(x, 0) = 0 \].Since \(f(0,0)=0\), the function is continuous in terms of \(x\).

Step 4: Evaluating \(f(0,y)\) for Continuity in \(y\) at Zero

Take the function for \(x = 0\):\[ f(0, y) = \frac{0 \cdot y^2}{0^4 + y^4} = 0 \].As \( y \to 0 \), \[ \lim_{{y \to 0}} f(0, y) = 0 \]. Since \(f(0,0)=0\), the function is continuous in terms of \(y\).

Step 5: Part B: Assessing Continuity Jointly at \((0,0)\)

To check continuity at \((0,0)\), examine the limit of \(f(x, y)\) as the point \((x, y)\) approaches \((0, 0)\) from various paths, like along x-axis \(y=mx\), y-axis \(x=0\), or any arbitrary path.

Step 6: Path Analysis for Discontinuity at \((0,0)\)

Check by approaching along the line \(y = x\): \[ f(x, x) = \frac{x \cdot x^2}{x^4 + x^4} = \frac{x^3}{2x^4} = \frac{1}{2x} \].As \(x \to 0\), \(\frac{1}{2x} \to \infty\), which does not approach \(0\).This is different from what is expected for continuity (which would be \(0\)). Hence, \(f\) is not continuous at \((0,0)\).

Key Concepts

Continuity

In mathematics, continuity of a function essentially means that the graph of the function is unbroken at a particular point or over an interval. For multivariable functions, like our function \(f(x, y)\), we need to assess continuity in terms of each variable separately and jointly. This means observing the behavior of the function as each variable separately approaches its limit.

In part (a) of the original exercise, we analyzed the function separately in terms of \(x\) and \(y\). We checked for the values when either variable was held constant while the other approached zero. This separate continuity is verified by ensuring that as \(x\) approaches 0 or \(y\) approaches 0, the function value approaches the same limit which matches the function's value at that point. Here, for both functions \(f(x, 0)\) and \(f(0, y)\), we find they equal 0 when \(x=0\) or \(y=0\). Thus, implying the function is continuous separately in each variable.

Ensuring a multivariable function is continuous separately is a starting point. However, it does not confirm full continuity at a point, as it needs to be checked through multiple paths.

Discontinuity

Discontinuity in a function occurs when there is a break or jump at a particular point, meaning the left-hand and right-hand limits at that point do not agree, or the limit does not equal the function's value. For multivariable functions, the concept of continuity can become trickier, as we must consider paths of approach.

In part (b) of the exercise, the function \(f(x, y)\) is examined as the point \((x, y)\) approaches the origin \((0,0)\). Although we found the function continuous in terms of each variable separately, examining different paths towards \((0,0)\) illustrates discontinuity. As highlighted in the solution, approaching the origin along the path \(y = x\) results in a limit \(\frac{1}{2x}\) that tends towards infinity as \(x\) approaches 0. Since this result does not match the expected \(f(0,0) = 0\), it indicates discontinuity.

This difference highlights the need to check all possible paths of approach when confirming the continuity of multivariable functions. Simply observing continuity in individual variables can overlook certain paths that lead to different outcomes.

Multivariable Functions

Multivariable functions, unlike single-variable functions, depend on more than one input variable. In their simplest form, they are defined by equations with more than one variable, such as \(f(x, y)\).

These functions are fundamentally different in how they present continuity. For instance, a multivariable function might seem continuous along individual axes or directions, but that does not confirm continuity everywhere in the plane or space it is defined, such as at a specific point like \((0,0)\) in this exercise.

With multivariable functions, paths towards a point are infinitely diverse and must be considered in assessing continuity or discontinuity at a point. The exercise highlighted that despite the function appearing well-behaved along individual axes, alternative paths showed a different behavior. This is a crucial aspect of learning and working with multivariable calculus.

• Consider individual paths of approach.
• Understand that apparent individual continuity does not guarantee full continuity.
• Scrutinize diverse paths, especially when suspecting possible discrepancies.