Chapter 11, Problem 47RE

To determine

To graph:

The limit value of the following functions:

  $\underset{x\to -2}{\lim }f\left(x\right),f\left(x\right)=\{\begin{array}{l}4-x,x\le -2\\ x^2-3,x>-2\end{array}$

Explanation of Solution

Given information:

  $\underset{x\to -2}{\lim }f\left(x\right),f\left(x\right)=\{\begin{array}{l}4-x,x\le -2\\ x^2-3,x>-2\end{array}$

Graph:

  

Graph 1: $\underset{x\to -2}{\lim }\left(4-x\right)$

  Graph 2: $\underset{x\to -2}{\lim }\left(x^2-3\right)$

Interpretation:

Know that:

When the limit value of $\underset{x\to a-}{\lim }f\left(x\right)$ and $\underset{x\to a+}{\lim }f\left(x\right)$ is not same then the function does not exist any limit value.

  $\underset{x\to a-}{\lim }f\left(x\right)\ne \underset{x\to a+}{\lim }f\left(x\right)$

From the

The given function $\underset{x\to -2}{\lim }f\left(x\right)=\{\begin{array}{l}4-x,x\le -2\\ x^2-3,x>-2\end{array}$ is graphed as above.

From the graph:

The function reaches $6$ and $1$ as it approaches $-2$ from right and left side respectively.

The on sided limit of the function as $x\to -2$ is calculated as:

  $\begin{array}{l}\underset{x\to -2}{\lim }f\left(x\right)=\underset{x\to -2}{\lim }4-x=4+2=6\\ \underset{x\to -2}{\lim }f\left(x\right)=\underset{x\to -2}{\lim }{x}^2-3={\left(-2\right)}^2-3=1\\ \underset{x\to -2}{\lim }4-x\ne \underset{x\to -2}{\lim }{x}^2-3\end{array}$

Hence, there is no limit value for the function $\underset{x\to -2}{\lim }f\left(x\right)=\{\begin{array}{l}4-x,x\le -2\\ x^2-3,x>-2\end{array}$ .