Chapter 11, Problem 2RE

Solution

a)

To calculate: The value of the limit $\underset{x\to 1^{-1}}{\lim }f\left(x\right)$ using the given graph.

The left-hand limit of the given function at $x=1$ is $-1$ .

Given information:

The graph is given below.

  

The left-hand limit of a function at a point is defined as the approaching value of the function from the left side at that particular point.

From the provided graph, it can be observed that the function approaches $-1$ from left side of $x=1$ . This implies that,

  $\begin{array}{c}\text{LHL}=\underset{x\to 1^{-1}}{\lim }f\left(x\right)\\ =2\end{array}$

Thus, the left-hand limit of the given function at $x=1$ is $-1$ .

b)

To calculate: The value of the limit $\underset{x\to 1}{\lim }f\left(x\right)$ using the given graph.

The limit of the given function at $x=1$ dos not exist.

Given information:

The graph is given below.

  

The left-hand limit of a function at a point is defined as the approaching value of the function from the left side at that particular point.

From the provided graph, it can be observed that the function approaches $-1$ from left side of $x=1$ . This implies that,

  $\begin{array}{c}\text{LHL}=\underset{x\to 1^{-}}{\lim }f\left(x\right)\\ =-1\end{array}$

The right-hand limit of a function at a point is defined as the approaching value of the function from the right side at that particular point.

From the provided graph, it can be observed that the function approaches 1 from right side of $x=1$ . This implies that,

  $\begin{array}{c}\text{RHL}=\underset{x\to 1^{+}}{\lim }f\left(x\right)\\ =1\end{array}$

It is known that the limit of a function at a point exist if left-hand limit and right limit at that point are equal. From the above calculations, it can be concluded that $\text{LHL}\ne \text{RHL}$ .

Thus, the limit of the given function at $x=1$ does not exist.