Chapter 11.1, Problem 30E

To determine

To graph:

The following functions and the value of limit:

  $f\left(x\right)=\{\begin{array}{l}2x-3,x<2\\ x^2-4x+5,x\ge 2\end{array}$

Explanation of Solution

Given information:

  $f\left(x\right)=\{\begin{array}{l}2x-3,x<2\\ x^2-4x+5,x\ge 2\end{array}$

Graph:

  

Graph 1: $\underset{x\to 2}{\lim }2x-3$

  

Graph 2: $\underset{x\to 2}{\lim }{x}^2-4x+5$

Interpretation:

There are two functions given as:

  $f\left(x\right)=\{\begin{array}{l}2x-3,x<2\\ x^2-4x+5,x\ge 2\end{array}$

The graphs of limits are graphed as above. The limit value for two functions is calculated as:

  $\begin{array}{l}\underset{x\to 2}{\lim }2x-3=2\left(2\right)-3=1\\ \underset{x\to 2}{\lim }{x}^2-4x+5={\left(2\right)}^2-4\left(2\right)+5=1\end{array}$

The limit value for both functions is $1$ .