Q 11. - Calculus

Question

Review the definition of continuity of a function of a single variable, $f$, at a point. Why is it necessary for $f$ to be defined on the union of two open intervals? 

Step-by-Step Solution

Answer

The function itself may or may not pass through the point. It is possible that the function may be indeterminate at this point. Thus, the limit evaluates the output value to which the function approaches as the input approaches the given point.

Step 1: Given Information

$f$ is a single-variable function that is known to be continuous at a given point. The goal is to demonstrate why the function $f$ must be defined on the union of two open intervals.

Step 2: Defining the limit

The function is described as a single variable function. Assume $x$ is the input variable, and the function is $f\left(x\right)$. Because the function has only one variable, it is $R$.

This function's graph will contain two variables. As a result, it will be in $R^2$.

Step 3: Evaluating the limit

When the limiting value of a function equals the value of the function at a given point, it is said to be continuous. Assume that the limiting value is L and that the point is $x=a$.

Write the function's limit.

$\underset{\underset{}{\left(x\right)\to \left(a\right)}}{\lim }f\left(x\right)=L=f\left(a\right)$

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 number $\delta >0$ is a real number