Question

In Exercises 31 and \(32,\) discuss the continuity of the functions \(f\) and \(g .\) Explain any differences. $$ \begin{array}{ll} f(x, y)=\left\\{\begin{array}{ll} \frac{4 x^{2} y^{2}}{x^{2}+y^{2}}, & (x, y) & \neq(0,0) \\ 0, & (x, y) & =(0,0) \end{array}\right. \\ g(x, y)=\left\\{\begin{array}{ll} \frac{4 x^{2} y^{2}}{x^{2}+y^{2}}, & (x, y) \neq(0,0) \\ 2, & (x, y)=(0,0) \end{array}\right. \end{array} $$

Short answer

The function \(f\) is continuous at the point (0,0) but the function \(g\) is not continuous at the point (0,0).

Step-by-step solution

Analyze the function \(f\)

We first find the limit of this function as (x,y) approaches (0,0). \[ \lim_{(x,y) \to (0,0)} \frac{4 x^{2} y^{2}}{x^{2}+y^{2}} \] For this, let's take y = mx and substitute it into the function \(f(x, mx)=\frac{4 x^{2} m^{2} x^{2}}{x^{2}+m^{2} x^{2}} = \frac{4 m^{2} x^{4}}{x^{2}(1+m^{2})} = \frac{4 m^{2} x^{2}}{1+m^{2}}\). Taking the limit as x approaches 0, we get \[ \lim_{x \to 0} \frac{4 m^{2} x^{2}}{1+m^{2}} = 0 \] This limit is independent of the slope m, therefore the limit as (x,y) approaches (0,0) exists. Since the function \(f\) is defined at (0,0) and the limit at that point exists and equals to function's value at that point, the function \(f\) is continuous at the origin.

Analyze the function \(g\)

Now let's do the same for the function \(g\). The function \(g\) is identical to \(f\) everywhere except at the origin, where \(g\) is defined to be 2. So, we find the limit of \(g\) as (x,y) approaches (0,0) as before: \[ \lim_{(x,y) \to (0,0)} \frac{4 x^{2} y^{2}}{x^{2}+y^{2}} \] Substituting y = mx and calculating the limit as before, we get \[ \lim_{x \to 0} \frac{4 m^{2} x^{2}}{1+m^{2}} = 0 \] However, the function \(g\) is defined to be 2 at the point (0,0), not 0. Therefore, the limit as (x,y) approaches (0,0) does not equal function's value at that point, and thus \(g\) is not continuous at the origin.