discuss the continuity of the function. continuous except where x^2 + y^2 < 1 continuous except at (0, 0)…

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

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

Answer

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))