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Chapter 12: Problem 52

At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$f(x, y)=\left\\{\begin{array}{ll} \frac{1-\cos \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}} & \text { if }(x, y) \neq(0,0) \\ 0 & \text { if }(x, y)=(0,0) \end{array}\right.$$

Short Answer

Expert verified
In this exercise, we were asked to find the points in the plane \(\mathbb{R}^2\) where the given function is continuous. We first established that the function is continuous for all points \((x, y) \neq (0,0)\) since it is a rational function, which is continuous everywhere except where the denominator is equal to zero. We then showed that the function is also continuous at \((0,0)\) by calculating the limit as \((x,y)\) approaches \((0,0)\) and verifying that it is equal to the function value at \((0,0)\). Thus, the function is continuous at every point in \(\mathbb{R}^2\).

Step by step solution

01

Continuity of \(f(x, y)\) for \((x, y) \neq (0,0)\)

Since the function \(f(x, y)\) is defined by a rational function (a quotient of two continuous functions) over the domain \((x, y) \neq (0,0)\), we can conclude that \(f(x, y)\) is continuous for all points \((x, y) \neq (0,0)\). The rational function is continuous everywhere except where the denominator is equal to zero. In this case, the denominator is \(x^2+y^2\), so the only point where it is equal to zero is \((0,0)\). Thus, \(f(x, y)\) is continuous at every point in \(\mathbb{R}^2\) except \((0,0)\).
02

Analyzing continuity at \((0,0)\)

Now, we will analyze the continuity of \(f(x, y)\) at the point \((0,0)\). The function \(f(x, y)\) is continuous at \((0,0)\) if the following limit exists and is equal to the function value at \((0,0)\), which is \(0\) in this case: $$\lim_{(x,y) \to (0,0)} \frac{1-\cos\left(x^2+y^2\right)}{x^2+y^2}.$$
03

Evaluating the limit using polar coordinates

To evaluate the limit as \((x, y) \to (0,0)\), we will use polar coordinates \(x = r\cos\theta\) and \(y = r\sin\theta\), where \(r \ge 0\) and \(0 \le \theta < 2\pi\). Then, the limit can be rewritten as: $$\lim_{r \to 0} \frac{1 - \cos\left(r^2\left(\cos^2\theta + \sin^2\theta\right)\right)}{r^2\left(\cos^2\theta + \sin^2\theta\right)}.$$ Using the identity \(\cos^2\theta + \sin^2\theta = 1\), the limit becomes: $$\lim_{r \to 0} \frac{1-\cos\left(r^2\right)}{r^2}.$$ Now, to check if the limit exists, we can use L'Hopital's rule in this form since \(\cos(0)=1\) and \(0/0\). Taking the derivative with respect to \(r\) of both numerator and denominator, we have: $$\lim_{r \to 0} \frac{-2r\sin\left(r^2\right)}{2r}.$$ Now we can cancel out the common factor of \(2r\), yielding: $$\lim_{r \to 0} -\sin\left(r^2\right).$$ As \(r\) approaches \(0\), \(r^2\) also approaches \(0\). Therefore, we have: $$\lim_{r \to 0} -\sin\left(r^2\right) = -\sin(0) = 0.$$ Thus, the limit exists and it is equal to the value of the function at \((0,0)\), which is \(0\).
04

Conclusion

Since we have shown that the limit exists at \((0,0)\) and is equal to the function value at \((0,0)\), we can conclude that \(f(x, y)\) is continuous at \((0,0)\). As a result, \(f(x, y)\) is continuous at every point in \(\mathbb{R}^2\).

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