Question

In Exercises 71 and \(72,\) use spherical coordinates to find the limit. [Hint: Let \(x=\rho \sin \phi \cos \theta, \quad y=\rho \sin \phi \sin \theta,\) and \(z=\rho \cos \phi,\) and note that \((x, y, z) \rightarrow(0,0,0)\) implies \(\left.\rho \rightarrow 0^{+} .\right]\) \(\lim _{(x, y, z) \rightarrow(0,0,0)} \frac{x y z}{x^{2}+y^{2}+z^{2}}\)

Short answer

The limit of the given function as \((x, y, z)\) approaches \((0, 0, 0)\) in spherical coordinates is \(0\).

Step-by-step solution

Transform to spherical coordinates

First, substitute the given transformations for \(x, y, z\) into spherical coordinates to the original function. Thus after substitution the given function becomes \[\frac{\rho^{3}\sin^{2}{\phi}\sin{\phi}\cos{\theta}\sin{\theta}}{\rho^{2}}\] which simplifies to \(\rho\sin^{3}{\phi}\cos{\theta}\sin{\theta}\)

Analyze under the limit

Considering that \(\rho \rightarrow 0^{+}\), it can be inferred that as \(\rho\) approaches 0, so does the entire expression. Hence, we get that \(\lim_{(x, y, z) \rightarrow(0,0,0)} \frac{x y z}{x^{2}+y^{2}+z^{2}}\) after transformation into spherical coordinates is \(0\).

Final Summary

The original limit in Cartesian coordinates was found by manipulating the variables \(x, y, z\) into spherical coordinates and then applying the limit to the new expression.