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 \(\left.\rho \rightarrow 0^{+} .\right]\) \(\lim _{(x, y, z) \rightarrow(0,0,0)} \tan ^{-1}\left[\frac{1}{x^{2}+y^{2}+z^{2}}\right]\)

Short answer

The limit as (x, y, z) tends towards (0,0,0) is \(\frac{\pi}{2}\)

Step-by-step solution

Convert Cartesian to Spherical Coordinates

Use the given equations for conversion. Replace \(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta,\) and \(\rho \cos \phi,\) with x, y, and z respectively in the function to be evaluated.

Simplify the function

After replacing, the function \(\tan ^{-1}\left[\frac{1}{x^{2}+y^{2}+z^{2}}\right]\) simplifies to \(\tan ^{-1}\left[\frac{1}{\rho^{2}}\right]\)

Evaluate Limit

Now that the function is simplified, evaluate the limit as \(\rho\) tends towards 0+. This is straightforward because when \(\rho\) approaches 0+, \(\frac{1}{\rho^{2}}\) tends towards infinity. The arctan of infinity is \(\frac{\pi}{2}\). So, the limit as the point approaches (0,0,0) is \(\frac{\pi}{2}\)