Question

discuss the continuity of the function. continuous except where x^2 + y^2 < 1 continuous except at (0, 0) continuous except where x^2 + y^2 > 1 continuous everywhere continuous except where x^2 + y^2 > 0 evaluate the limit of f(x, y) (if it exists) as (x, y)→(0, 0). (if an answer does not exist, enter dne.) f(x, y)=1 - \frac{cos(x^2 + y^2)}{x^2 + y^2} graph graph description

Explanation:

Step1: Use polar - coordinates

Let \(x = r\cos\theta\) and \(y = r\sin\theta\), then \(x^{2}+y^{2}=r^{2}\), and as \((x,y)\to(0,0)\), \(r\to0\). The function \(f(x,y)=\frac{1 - \cos(x^{2}+y^{2})}{x^{2}+y^{2}}\) becomes \(f(r,\theta)=\frac{1-\cos(r^{2})}{r^{2}}\).

Step2: Apply the double - angle formula or the well - known limit

We know the well - known limit \(\lim_{u\to0}\frac{1 - \cos u}{u}=\ 0\). Let \(u = r^{2}\), as \(r\to0\), \(u\to0\). So \(\lim_{r\to0}\frac{1-\cos(r^{2})}{r^{2}}=\lim_{u\to0}\frac{1 - \cos u}{u}\times\lim_{r\to0}r = 0\).

Step3: Analyze continuity

A function \(z = f(x,y)\) is continuous at \((a,b)\) if \(\lim_{(x,y)\to(a,b)}f(x,y)=f(a,b)\). The function \(f(x,y)=\frac{1 - \cos(x^{2}+y^{2})}{x^{2}+y^{2}}\) is not defined at \((0,0)\) since we have a \(\frac{0}{0}\) form at \((0,0)\), but \(\lim_{(x,y)\to(0,0)}f(x,y) = 0\). The function is continuous everywhere except at \((0,0)\).

Answer:

continuous except at \((0,0)\)