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 because the limit at (0,0,0) depends on the path taken to approach it.

Step 1: Understand Continuity Criterion

A function \( f(x, y, z) \) is continuous at a point \((a, b, c)\) if \[ \lim_{(x, y, z) \to (a, b, c)} f(x, y, z) = f(a, b, c). \] In our case, since \( f(0, 0, 0) = 0 \), we need to show that \[ \lim_{(x, y, z) \to (0, 0, 0)} f(x, y, z) eq 0. \] If the limit does not equal zero, the function is not continuous at \((0, 0, 0)\).

Step 2: Evaluate Using the Path \((x, y, z) = (t, 0, t)\)

Substitute \( x = t \), \( y = 0 \), and \( z = t \) into \( f(x, y, z) \): \[ f(t, 0, t) = (0+1) \frac{t^{2} - t^{2}}{t^{2} + t^{2}} = \frac{0}{2t^{2}} = 0. \] Thus, along this path, \( \lim_{t \to 0} f(t, 0, t) = 0 \).

Step 3: Evaluate Using the Path \((x, y, z) = (t, t, 0)\)

Now, substitute \( x = t \), \( y = t \), and \( z = 0 \) into \( f(x, y, z) \): \[ f(t, t, 0) = (t+1) \frac{t^{2} - 0^{2}}{t^{2} + 0^{2}} = (t+1). \] Thus, along this path, \( \lim_{t \to 0} f(t, t, 0) = 1 \).

Step 4: Compare Path Results

The two limits obtained from different paths are \(0\) and \(1\), indicating a contradiction. Since the limit depends on the chosen path, \( \lim_{(x, y, z) \to (0, 0, 0)} f(x, y, z) \) does not exist, proving that \( f(x, y, z) \) is not continuous at \( (0, 0, 0) \).

Key Concepts

Continuity in 3D

When discussing continuity in 3D, we are dealing with functions that have three variables, usually written as \( f(x, y, z) \). For such a function to be continuous at a particular point \((a, b, c)\), the limit of the function as the point \((x, y, z)\) moves towards \((a, b, c)\) must be equal to the value of the function at \((a, b, c)\).
This can be expressed as:

• If \( \lim_{(x, y, z) \to (a, b, c)} f(x, y, z) = f(a, b, c) \), then the function is continuous at that point.
• Hence, continuity ensures a function shows no sudden jumps or breaks at the point concerned.

With this in mind, for the given exercise, we need to check if the limit of \( f(x, y, z) \) as \((x, y, z)\) approaches \((0, 0, 0)\) equals \( f(0, 0, 0) \), which is \( 0 \). If it doesn’t match, the function is not continuous.

Path-dependent limits

Path-dependent limits occur when the limit of a function as it approaches a point from different directions or "paths" varies. In multivariable calculus, this phenomenon is crucial because it can signal discontinuities.
For a 3D function \( f(x, y, z) \), if the limits differ based on the chosen path of approach, the overall limit does not exist at that point.

• For example, if taking the path \((x, y, z) = (t, 0, t)\), the evaluated limit for the given function is \( 0 \).
• On a different path, such as \((x, y, z) = (t, t, 0)\), the limit turns out to be \( 1 \).

This conflict in values shows that the limit depends on the path taken and solidifies the claim that the function can’t be continuous at \((0, 0, 0)\). For continuity, consistent limit values are essential, irrespective of the approach path.

Limit evaluation in multiple dimensions

Evaluating limits in multiple dimensions is more nuanced than in single-dimensional cases. In 1D, limits depend solely on approaching a point from the left or right on a number line. However, in 3D, it can approach from countless directions, adding complexity.
To accurately evaluate a limit, one must consider all possible paths or directions of approach.

• It is crucial to test different paths to ensure consistency in limit values.
• If limits vary based on paths, then a unique limit does not exist at that point in 3D space.

In the given function example, different paths toward \((0, 0, 0)\) clearly lead to different limits. Hence, the challenge of limit evaluation in multiple dimensions is essential to recognizing when continuity fails due to path-dependent behavior.