Problem 73 - Calculus Early Transcendentals

Question

Let $$f(x, y)=\left\\{\begin{array}{ll}\frac{\sin \left(x^{2}+y^{2}-1\right)}{x^{2}+y^{2}-1} & \text { if } x^{2}+y^{2} \neq 1 \\\b & \text { if } x^{2}+y^{2}=1\end{array}\right.$$ Find the value of \(b\) for which \(f\) is continuous at all points in \(\mathbb{R}^{2}\).

Step-by-Step Solution

Answer

For the function \(f(x, y)\) to be continuous at all points in \(\mathbb{R}^2\), the value of \(b\) must be 1.

Step 1: Determine the continuity of the function when \(x^2 + y^2 \neq 1\)

First, let's consider the function \(f(x, y)\) in the region where \(x^2 + y^2 \neq 1\). Here, the function is given by: $$ f(x, y) = \frac{\sin (x^2 + y^2 - 1)}{x^2 + y^2 - 1}. $$ As we approach the origin \((0,0)\), we can check the limit to ensure continuity: $$ \lim_{(x,y) \to (0,0)} \frac{\sin (x^2 + y^2 - 1)}{x^2 + y^2 - 1}. $$ Using the well-known limit \(\lim_{z \to 0} \frac{\sin z}{z} = 1\), we can rewrite the limit as: $$ \lim_{(x,y) \to (0,0)} \frac{\sin (x^2 + y^2 - 1)}{x^2 + y^2 - 1} = \lim_{z \to 0} \frac{\sin z}{z} = 1, $$ where \(z = x^2 + y^2 - 1\). Therefore, the function is continuous when \(x^2 + y^2 \neq 1\).

Step 2: Find the value of \(b\) for which \(f\) is continuous at \(x^2 + y^2 = 1\)

Now we need to find the value of \(b\) for which the function is continuous at the points where \(x^2 + y^2 = 1\). For this continuity, we require that the limit of the function as we approach these points is equal to \(b\). So, we need to evaluate: $$ \lim_{x^2 + y^2 \to 1} \frac{\sin (x^2 + y^2 - 1)}{x^2 + y^2 - 1}. $$ Substituting \(z = x^2 + y^2 - 1\) as before, we can evaluate the limit as \(z \to 0\): $$ \lim_{z \to 0} \frac{\sin z}{z} = 1. $$ Therefore, for the function to be continuous at all points in \(\mathbb{R}^2\), we need \(b = 1\).

Key Concepts

Limits in Multivariable Calculus

In multivariable calculus, understanding limits involves exploring how a function behaves as its input variables approach a certain point. Different from single-variable calculus, here you have two or more variables approaching a point simultaneously.
This can make visualizing limits more challenging. In the given exercise, we deal with a function of two variables, \(x\) and \(y\), defined by the piecewise function \(f(x,y)\). To check the continuity of the function in the region where \(x^2 + y^2 eq 1\), we calculated:- The limit as \((x,y)\) approaches \((0,0)\), which rewrote to a simpler form using \(z = x^2 + y^2 - 1\).
- This simplification helped us apply the basic limit \(\lim_{z \to 0} \frac{\sin z}{z} = 1\).
This limit demonstrates that the function's value approaches a certain number consistently from all paths, indicating continuity, except at the point \(x^2 + y^2 = 1\). Remember, checking limits from multiple paths is crucial in multivariable calculus because approaching from different angles might reveal potential discontinuities.

Piecewise Functions

Piecewise functions are those that have different expressions for different parts of their domain. They are particularly handy for describing complex scenarios where a single formula isn’t enough.
Such a function might change its behavior based on the input values, much like our function \(f(x, y)\) changes depending on whether \(x^2 + y^2\) equals 1 or not.Understanding how to handle these functions:- If \(x^2 + y^2 eq 1\), we use \(\frac{\sin(x^2 + y^2 - 1)}{x^2 + y^2 - 1}\) as the function output.
- If \(x^2 + y^2 = 1\), the function outputs a preset value \(b\).
The trick is to choose \(b\) so that the function becomes continuous everywhere. Continuity at these piece points requires matching the function's limit, as \((x, y)\) goes to the piece point, with the function value defined there. In this case, since the limit we calculated was 1, setting \(b = 1\) ensures seamless continuity, precisely illustrating how a piecewise function can be unified across its domain.

Trigonomeric Limits

Trigonometric limits often appear in calculus because trigonometric functions like sine and cosine have specific behaviors near critical points, such as zero.
These limits are incredibly useful in simplifying complex expressions involving trigonometric functions. In the problem at hand, we made use of the well-known trigonometric limit:- \(\lim_{z \to 0} \frac{\sin z}{z} = 1\), a fundamental result.
This tells us that - As \(z\) approaches zero, the ratio of \(\sin z\) to \(z\) approaches 1.
This concept allowed us to resolve the indeterminacy presented when \(x^2 + y^2 \) is nearly equal to 1 by simplifying the fraction involved in the function \(f(x, y)\).
It's a beautiful example of how trigonometric limits can resolve potential discontinuities, essentially making the seemingly complex behavior of trigonometric functions manageable, especially in multivariable settings.