Q 7. - Calculus

Question

How does the definition of the limit of a function of a single variable, $f$, imply that $f$is defined on the union of two open intervals? 

Step-by-Step Solution

Answer

It is implied the function will exist and can be written as $(a-\delta ,a)\cup (a,a+\delta )$, where $\delta >0$is a real number.

Step 1: Given Information

The limit of a function of single variable 'f'  is defined. The objective is to explain how does it imply that 'f' is defined on the union of two open intervals. 

Step 2: Defining the limit

The function $f$ is said to be in single variable. Take the input variable as 'x', thus the function is$f\left(x\right)$. The function is single variable and hence it is in $R$. This function's graph will contain two variables. Hence, it will be in $R^2$.

Step 3: Evaluating the limit

The function's limit is said to be defined at a specific point. Assume that the point is $r=a$ and that the limiting value is L. Where is the functional limit:$\underset{\left(x\right)\to \left(a\right)}{\lim }f\left(x\right)=L$

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 function may or may not exist at the specified point, but it will exist for the route before and after this point, as shown above.

As a result, the interval in which the function will exist is $(a-\delta ,a)\cup (a,a+\delta )$. where the $\delta >0$ is a real number.