Q 8. - Calculus

Question

How does the definition of the limit of a function of two variables, $f$ , imply that  $f$is defined on an open subset of $R^2$? 

Step-by-Step Solution

Answer

It can be implied that the function will exist for the interval : $0<\sqrt{{(x-a)}^2+{(y-b)}^2}<\delta$,where $\delta >0$is a real number.

Step 1: Given Information

Consider the case where the limit of a two-variable function f is known. The goal is to show what it means when f is defined as the union of two open intervals.

Step 2: Defining the limit

$f$ is stated to be a two-variable function.

Assume the input variables are $x,y$, making the function $f(x,y)$. Because the function has two variables, it is in $R^2$.

This function's graph will contain three variables. Hence, it will be in$R^3$.

Step 3: Evaluating the limit

The limit of function is said to be defined at a point. Assume the point to be $(x,y)=(a,b)$and assume the limiting value to be L. The limit of function is $\underset{(x,y)\to (a,b)}{\lim }f(x,y)=L$ 

Step 4: Limiting value of function

The output that the function tends to approach as the input approaches the point is determined by evaluating the function's limit at a point. The path of approach is irrelevant for determining the limit unless it passes via the input location. As a result, regardless of the approach path, the limiting value will remain constant. The function may or may not travel through the point as well. It's conceivable that the function is now uncertain.

As the input approaches the designated point, the limit evaluates the output value to which the function approaches.

Step 5: Existence of the limit.

The above explanation makes it clear that the function may or may not exist at the given point, but it will exist for the open disk around this point. Thus, the interval for which the function will exist can be written as $0<\sqrt{{(x-a)}^2+{(y-b)}^2}<\delta$ ,where $\delta >0$is a real number.