Q 9. - Calculus

Question

How does the definition of the limit of a function of three variables,$f$, imply that $f$is defined on an open subset of $R^3$? 

Step-by-Step Solution

Answer

It is implied the function will exist and can be written as $0<\sqrt{{(x-a)}^2+{(y-b)}^2+{(z-c)}^2}<\delta$, where $\delta >0$is a real number.

Step 1: Given Information

Consider that the limit of a function of three variables $f$is defined. The objective is to explain how does it imply that $f$ is defined on the open subset of $R^3$

Step 2: Defining the limit

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

Assume $x,y,z$ are the input variables, and the function is $f(x,y,z)$. The function has three variables, hence it is in $R^3$. This function's graph will have four variables. As a result, it will be in $R^4$.

Step 3: Evaluating the limit

Consider that the function's limit is said to be defined at a specific point $(a,b,c)$

Assume that the limiting value is L and that the point is $(x,y,z) =(a,b,c)$.

The function's limit is $\underset{(x,y,z)\to (a,b,c)}{\lim }f(x,y,z)=L$.

Step 4: Limiting value of function

The function's limit is 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 function may or may not exist at the specified point, but it will exist for the open ball surrounding this point, as explained above. As a result, the interval during which the function will exist is $0<\sqrt{{(x-a)}^2+{(y-b)}^2+{(z-c)}^2}<\delta$,where $\delta >0$ is a real number.