Question

discuss the continuity of the function. f(x, y, z) = sin(z) / (e^x + e^y) continuous everywhere continuous except where x < 0 continuous except where y < 0 continuous except where z < 0 continuous except at (0, 0, 0)

Explanation:

Step1: Recall continuity condition

A function \(f(x,y,z)=\frac{\sin(z)}{e^{x}+e^{y}}\) is continuous at a point \((a,b,c)\) if \(\lim_{(x,y,z)\to(a,b,c)}f(x,y,z)=f(a,b,c)\) and the function is well - defined at \((a,b,c)\). The exponential functions \(y = e^{x}\) and \(y=e^{y}\) are defined for all real \(x\) and \(y\), and the sine function \(y = \sin(z)\) is defined for all real \(z\). However, the denominator \(e^{x}+e^{y}
eq0\) for all real \(x\) and \(y\) since \(e^{x}>0\) and \(e^{y}>0\) for all \(x,y\in R\). The function \(f(x,y,z)\) is undefined when \(\sin(z) = 0\), i.e., \(z = k\pi\), \(k\in\mathbb{Z}\).

Step2: Analyze the domain of continuity

The function \(f(x,y,z)\) is a quotient of two continuous functions. The numerator \(g(z)=\sin(z)\) and the denominator \(h(x,y)=e^{x}+e^{y}\). The function \(f(x,y,z)\) is continuous everywhere except where \(\sin(z)=0\), or \(z = k\pi,k\in\mathbb{Z}\). When \(z = k\pi\), the denominator \(e^{x}+e^{y}
eq0\) for all real \(x\) and \(y\), but the function is not well - defined due to the zero in the numerator.

Answer:

continuous except where \(z = k\pi,k\in\mathbb{Z}\) (none of the given options are correct as stated in the problem - if we assume the closest is the option about \(z\) values, we can say the function is continuous except where \(z<0\) is a wrong statement, continuous everywhere is wrong, continuous except where \(y < 0\) is wrong, continuous except at \((0,0,0)\) is wrong. The correct answer should be continuous except where \(z\) is an integer multiple of \(\pi\))