Chapter 2.1, Problem 75E

To determine

To find: The correct choice for the value of $\underset{x\to 1}{\lim }f\left(x\right)$ .

Answer to Problem 75E

The correct choice is (E) such that the value of $\underset{x\to 1}{\lim }f\left(x\right)$ does not exist.

Explanation of Solution

Given information: The function is $f\left(x\right)=\{\begin{array}{l}2-x,x\le 1\\ \frac{x}{2}+1,x>1\end{array}$ .

The function $f\left(x\right)$ corresponding to the left hand side of $x=1$ is $2-x$ .

Simplify the given limit.

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

So, the value of $\underset{x\to 1^{-}}{\lim }f\left(x\right)$ is $1$ .

The function $f\left(x\right)$ corresponding to the right hand side of $x=1$ is $\frac{x}{2}+1$ .

Simplify the given limit.

  $\begin{array}{c}\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }\left(\frac{x}{2}+1\right)\\ =\underset{x\to 1^{+}}{\lim }\left(\frac{x}{2}\right)+\underset{x\to 1^{+}}{\lim }1\\ =\frac{1}{2}+1\\ =\frac{3}{2}\end{array}$

So, the value of $\underset{x\to 1^{+}}{\lim }f\left(x\right)$ is $\frac{3}{2}$ .

As the left hand limit and right hand limit for $f\left(x\right)$ at $x=1$ are not equal.

Therefore, the correct choice is (E) such that the value of $\underset{x\to 1}{\lim }f\left(x\right)$ does not exist.