Problem 4 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 6 , discuss the continuity of \(f\). \(f(x, y)= \begin{cases}\frac{x+y}{x^{2}+y^{2}} & \text { if }(x, y) \neq(0,0) \\\ 0 & \text { if }(x, y)=(0,0)\end{cases}\)

Step-by-Step Solution

Answer

Not continuous at , as different paths yield different limits.

Step 1 - Understand the Function

The function given is a piecewise function: If , otherwise.

Step 2 - Check Continuity at

A function is continuous at a point if the limit exists at that point and equals the function's value. Check the limit of as : .

Step 3 - Calculate Limit Along Different Paths

To determine the limit, examine different paths approaching . consider : simplifies to . Then, consider : simplifies to .

Step 4 - Compare Limits from Different Paths

Compare limits from different paths: along along . Since the results are different, the limit doesn't exist at .

Step 5 - Conclusion on Continuity

Since the limit doesn't exist or isn't equal to the function's value at , it's not continuous at . At other points where , the function is just .

Key Concepts

limits along different paths

When analyzing piecewise functions, it's crucial to explore the limits along various paths. This helps us determine whether the limit of the function exists at particular points. In our exercise, the function is:oindent \(f(x, y)=\begin{cases}\frac{x+y}{x^{2}+y^{2}} & \text { if }(x, y) eq(0,0)\0 & \text { if }(x, y)=(0,0)\end{cases}\)To find the limit as \((x, y) \rightarrow (0,0)\), we examined different paths.

• First, we considered the path \(y = x\). When substituting \(y\) with \(x\), the function simplifies to \(\frac{x+x}{x^2+x^2} = \frac{2x}{2x^2} = \frac{1}{x}\). As \(x\) approaches 0, this limit tends to infinity.
• Next, we considered the path \(y = 0\). When \(y = 0\), the function becomes \(\frac{x+0}{x^2+0} = \frac{x}{x^2} = \frac{1}{x}\). As \(x\) approaches 0, this also tends to infinity.

The limits along different paths are the same in this example, but they do not converge to a single finite value. Thus, the limit does not exist at \((0,0)\).

continuity at a point

Continuity at a point means that the function should have no breaks, jumps, or holes at that specific point. In mathematical terms, a function \(f(x, y)\) is continuous at point \((a,b)\) if three conditions are fulfilled:

• \(f(a, b)\) is defined.
• The limit of \(f(x, y)\) as \((x, y)\) approaches \((a,b)\) exists.
• The function's value at that point equals the limit, i.e., \(\lim_{(x,y)\to(a,b)}f(x,y) = f(a,b)\) .

For our piecewise function: oindent \(f(x, y)=\begin{cases}\frac{x+y}{x^{2}+y^{2}} & \text { if }(x, y) eq(0,0)\0 & \text { if }(x, y)=(0,0)\end{cases}\)Let's check continuity at the point \((0,0)\):

• \(f(0, 0)\) is defined and equals to 0.
• However, the limit as \((x, y)\to(0,0)\) does not exist.

Because the limit does not exist at \((0,0)\), the function is not continuous at this point. At other points where \((x, y)eq(0,0)\), the function does not break or jump, ensuring continuity wherever \((x, y)eq(0,0)\).

piecewise function analysis

Piecewise functions have different expressions based on different conditions. Analyzing them involves:

• Understanding the boundaries where different pieces apply.
• Ensuring the expressions align seamlessly at those boundaries.

For the functionoindent\(f(x, y)=\begin{cases}\frac{x+y}{x^{2}+y^{2}} & \text { if }(x, y) eq(0,0)\0 & \text { if }(x, y)=(0,0)\end{cases}\), the analysis includes:

• Checking the behavior of \(\frac{x+y}{x^2 + y^2}\) as \((x,y)eq(0,0)\). This helps us understand the pattern away from the origin.

We found that the function oindent \(\frac{x+y}{x^2+y^2}\) becomes undefined at \((0,0)\). Hence, the piecewise definition accounts for this by setting \(f(x, y)\) to 0 exactly at \((0,0)\), making the function defined everywhere in its domain.

• Finally, the alignments at boundaries should guarantee no sudden jumps or breaks. If a consistent limit from all paths exists, then it smooths out transitions.

In our case, due to differing behaviors along multiple paths to \(0,0\), continuity fails at this point. However, beyond this specific point, the function remains well-behaved and continuous everywhere else.