Q. 20 - Calculus

Question

Let f  be a function of three variables that is continuous everywhere. 

(a) Explain why the function $\frac{f(x,y,z)}{x+y+z}$is continuous if and only if $x+y+z\ne 0$.

(b) Use Definition $12.15$ to explain why $\underset{\underset{}{(x,y,z)\to (a,b-(a+b\left)\right)}}{\lim }\frac{f(x,y,z)}{x+y+z}.$ does not exist for any pair $(a,b)$of real numbers. 

Step-by-Step Solution

Answer

Part (a): If $x+y+z=0$ then the limit becomes indeterminate.

Part (b): At this point the denominator becomes zero.

Part (a): Step 1: Given Information

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 $x+y+z\ne 0$ is true.

The continuity of functions, like the general rule of quotient of limits, asserts that if functions $f(x,y) and g(x,y)$ 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 $g(x,y)\ne 0$

Part(a): Step 2: Continuity of the limit

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 $0$.

$x+y+z\ne 0$

Part (b): Step 1: Existence of the limit

The goal is to show why for every real number a, $\underset{\underset{}{(x,y,z)\to (a,b-(a+b\left)\right)}}{\lim }\frac{f(x,y,z)}{x+y+z}$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.