Chapter 2.2, Problem 2E

Explain what it means to say that

$\underset{x\to 1^{-}}{\lim }f(x)=3\text{and}\underset{x\to 1^{-}}{\lim }f(x)=7$

In this situation is it possible that $\underset{x\to 1}{\lim }f(x)$ exists? Explain.

To determine

To explain: The meaning of $\underset{x\to 1^{-}}{\lim }f\left(x\right)=3$ and $\underset{x\to 1^{+}}{\lim }f\left(x\right)=7$ and the existence of $\underset{x\to 1}{\lim }f\left(x\right)$.

Explanation of Solution

The meaning of $\underset{x\to 1^{-}}{\lim }f\left(x\right)=3$ is, as the values of x gets more closer to 1 from the left that is $x<1$, the value of $f\left(x\right)$ arbitrarily approaches 3.

The meaning of $\underset{x\to 1^{+}}{\lim }f\left(x\right)=7$ is, as the values of x gets more closer to 1 from the right that is $x>1$, the value of $f\left(x\right)$ arbitrarily approaches 7.

The equation $\underset{x\to 1}{\lim }f\left(x\right)=L$ does not exist since the left-hand limit and right-hand limit are not the same.

Recall the definition of the limit of a function that the limit exists when the right hand limit and the left hand limit are the same.

Since, the left-hand limit is 3 and the right hand limit is 7, $\underset{x\to 1}{\lim }f\left(x\right)$ does not exist.