Chapter 1, Problem 18RE

To determine

To find: Whether the $\underset{x\to c}{\lim }f\left(x\right)$ exists or not, for which the function $f\left(x\right)$ is shown below.

  

Answer to Problem 18RE

The value of $\underset{x\to c}{\lim }f\left(x\right)$ doesn’t exist.

Explanation of Solution

Given:

The graph for the function $f\left(x\right)$ is shown below.

  

Concept used:

The limit of the function $g\left(x\right)$ exists at some point $x=a$ in domain of $g\left(x\right)$ , if the function has no holes, jumps or breaks at that point $x=a$ in graph of function.

Calculation:

From the function $f\left(x\right)$ , it can be interpreted that the domain of the function $f\left(x\right)$ is real. And, it can be interpreted from the graph that the function $f\left(x\right)$ is discontinuous because of hole at $x=c$ and it is also discontinuous $x=c$ because there is a jump in function at $x=c$ as seen from the graph. Therefore, the value of $\underset{x\to c}{\lim }f\left(x\right)$ is not defined and finite. $\mathrm{......}\left(1\right)$

Further, it can be interpreted from the graph that as $x\to c^{-}$ (from left), the value of $\underset{x\to c^{-}}{\lim }f\left(x\right)$ is defined and finite because there exists a hole as $x\to c^{-}$ . Therefore, $x\to c^{-}$ doesn’t exist. And as $x\to c^{+}$ (from right), the value of $\underset{x\to c^{+}}{\lim }f\left(x\right)$ is defined and finite because there exists a dark hole as $x\to c^{+}$ . Therefore, $x\to c^{+}$ exists and it is finite. But $\underset{x\to c^{-}}{\lim }f\left(x\right)\ne \underset{x\to c^{+}}{\lim }f\left(x\right)$ as explained from $\left(1\right)$ .

Conclusion:

Hence, the value of $\underset{x\to c}{\lim }f\left(x\right)$ doesn’t exist.