Problem 48 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that the function defined by $$ f(x, y, z)=(y+1) \frac{x^{2}-z^{2}}{x^{2}+z^{2}} \quad \text { for }(x, y, z) \neq(0,0,0) $$ and \(f(0,0,0)=0\) is not continuous at \((0,0,0)\).

Step-by-Step Solution

Answer

The function is not continuous at \((0, 0, 0)\) because the limit is path-dependent.

Step 1: Understand Continuity

A function \( f \) is continuous at a point \((a, b, c)\) if \( \lim_{(x, y, z) \to (a, b, c)} f(x, y, z) = f(a, b, c) \). Here, we need to check if \( \lim_{(x, y, z) \to (0, 0, 0)} f(x, y, z) = 0 \) holds.

Step 2: Simplify the Expression

Rewriting the function, we have \( f(x, y, z) = (y+1) \left(\frac{x^2 - z^2}{x^2 + z^2}\right) \). Our task is to check the limit as \((x, y, z)\) approaches \((0, 0, 0)\).

Step 3: Evaluate the Limit Along a Path

Choose the path where \( x = t \), \( y = 0 \), and \( z = t \). This simplifies \( f(t, 0, t) = (0+1) \left(\frac{t^2 - t^2}{t^2 + t^2}\right) = \frac{0}{2t^2} = 0 \).

Step 4: Evaluate the Limit Along Another Path

Now, choose another path such as \( x = t \), \( y = t \), and \( z = 0 \). This gives \( f(t, t, 0) = (t+1) \left(\frac{t^2}{t^2}\right) = t+1 \). As \( t \to 0 \), \( f(t, t, 0) \) approaches 1.

Step 5: Compare the Limits

From Step 3, the limit along \( (t, 0, t) \) is 0, and from Step 4, the limit along \( (t, t, 0) \) is 1. These different results imply that the limit does not exist as \((x, y, z)\) approaches \((0, 0, 0)\).

Step 6: Conclude on Continuity

Since the limits are not the same along different paths, \( \lim_{(x, y, z) \to (0, 0, 0)} f(x, y, z) eq f(0, 0, 0) \). Therefore, the function is not continuous at \((0, 0, 0)\).

Key Concepts

Continuity of Functions

Continuity of a function is a concept that translates the intuitive idea of a graph being "unbroken". In multivariable calculus, a function \( f(x, y, z) \) is continuous at a point \((a, b, c)\) if the limit of the function as it approaches the point is equal to the function's value at that point. This can be mathematically represented as:

• \( \lim_{(x, y, z) \to (a, b, c)} f(x, y, z) = f(a, b, c) \)

This criterion ensures that, as you get infinitely closer to \((a, b, c)\), the function \( f(x, y, z) \) behaves predictably by matching its output to its direct value at \((a, b, c)\). Any deviation from this behavior can indicate a discontinuity. In simpler terms, imagine a smooth surface without any jumps or breaks - that's what continuity implies.

For the function \( f(x, y, z) \) given in the exercise, we need to evaluate if the condition of continuity holds true at the specific point \((0, 0, 0)\). If it doesn't, the function is said to be discontinuous at that point.

Limit of Functions

The limit of a function determines how a function behaves as it approaches a specific point. In multivariable calculus, evaluating a limit can involve exploring how the function changes from many different directions around a point. This is because, unlike in single-variable calculus, you can approach the point from various paths.

• Mathematically, \( \lim_{(x, y, z) \to (a, b, c)} f(x, y, z) \) is about finding the value the function settles into as the point \((x, y, z)\) gets close to \((a, b, c)\).

To illustrate, in our exercise, we consider different paths to understand the behavior of the function as it nears \((0, 0, 0)\).

By analyzing the paths:

• \( x = t \), \( y = 0 \), \( z = t \) results in a limit of 0, showing one potential behavior of the function.
• \( x = t \), \( y = t \), \( z = 0 \) results in a limit of \( t+1 \), contradicting the other path as it results towards 1 as \( t \to 0 \).

Each of these paths shows different limiting behavior, which is crucial in determining the overall existence of a single, coherent limit.

Path-Dependent Limits

Path-dependent limits come into play when considering limits for functions of several variables. In problems where there are multiple variables, the limit along different paths approaching a point can yield different results. This concept is especially important when the overall limit does not exist due to differing behaviors along various paths.

• If you can find even two paths with different limit results, you prove that the limit does not exist.

In our exercise, we experience path-dependent behavior, with limits along different routes leading to different outcomes. For example:

• Along the path \( (x, y, z) = (t, 0, t) \), the limit is 0.
• Along the path \( (x, y, z) = (t, t, 0) \), the limit yields 1 as \( t \to 0 \).

These differing results demonstrate that the limit of the function \( f(x, y, z) \) does not exist as \((x, y, z)\) approaches \((0,0,0)\). Hence, because the paths give no consistent value, the function lacks continuity at that point. By understanding path-dependent limits, we gain essential insights into how multivariable functions behave at crucial points.