Q. 19 - Calculus

Question

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

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

(b) Use Definition $12.15$ to explain why $\underset{(x,y)\to (a,a)}{\lim }\frac{f(x,y)}{x-y}$ does not exist for any real number a. 

Step-by-Step Solution

Answer

Part (a): Because if $x=y$, then the function becomes indeterminate  

Part (b): For point $(a,a)$ the denominator becomes zero.

Part (a): Step 1: Given Information

A two-variable function f is known to be continuous everywhere.

The goal is to demonstrate why the function $\frac{f(x,y)}{x-y}$ is continuous only if and only if $x\ne y$

The continuity of functions 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 $\frac{f(x,y)}{x-y}$, we use this rule.

Everywhere, the function $f(x,y)$ is said to be continuous.

A polynomial function is the denominator function $(x-y)$. As a result, it is also consistent throughout.

The sole remaining criterion is that the denominator does not equal $0$.

$$\begin{gathered} x-y\ne 0 \\ x\ne y \end{gathered}$$

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

The goal is to show why for every real number a, $\underset{(x,y)\to (a,a)}{\lim }\frac{f(x,y)}{x-y}$ does not exist.

The denominator of the above function becomes $0$ at the point $(a,a)$. As a result, the function is indefinite. As a result, the limit doesn't exist at this point.