91Ó°ÊÓ

Consider the function f, which is a three-variable function that is continuous everywhere.

The goal is to show why the function $\frac{f(x,y,z)}{x+y+z}$ is continuous only if and only if $x+y+z≠0$ is true.

The continuity of functions, like the general rule of quotient of limits, asserts that if functions $f(x,y)andg(x,y)$ are continuous in a certain interval, then the quotient function $\frac{f(x,y)}{g(x,y)}$is likewise continuous in the same interval, if and only if $g(x,y)≠0$

For the function $\frac{f(x,y,z)}{x+y+z}$we apply the same rule.

The function $f(x,y,z)$is assumed to be continuous in all directions. $(x+y+z)$is a polynomial function in the denominator. As a result, it is also consistent throughout. The sole remaining criterion is that the denominator does not equal $0$.

$x+y+z≠0$

The goal is to show why for every real numbera, $\underset{\underset{}{(x,y,z)→(a,b-(a+b\left)\right)}}{\lim }\frac{f(x,y,z)}{x+y+z}$does not exist.

The denominator of the above function becomes 0 at the point$(a-b,(a+b\left)\right)$. As a result, the function is indefinite. As a result, there exist no limit for this point.