Chapter 2.2, Problem 7E

Sketch the graph of the function and use it to determine the values of a for which $\underset{x\to a}{\lim }$f(x) exists.

$f(x)=\{\begin{array}{l}1+x\text{if}x<-1\\ x^2\text{if}-1\le x<1\\ 2-x\text{if}\text{x}\ge \text{1}\end{array}$

To determine

To sketch: The graph of $f\left(x\right)$ and determine the value of a from the graph for which $\underset{x\to a}{\lim }f\left(x\right)$ exists.

Explanation of Solution

Given:

The function $f\left(x\right)=\{\begin{array}{ll}1+x\hfill & \text{if }x<-1\hfill \\ x^2\hfill & \text{if }-1\le x<1\hfill \\ 2-x\hfill & \text{if }x\ge 1\hfill \end{array}$.

Graph:

Draw a parabola of $y=x^2$ between the interval $-1\le x<1$ and draw a ray of $y=2-x$ starting at the point (1, 1) on a given interval $x\ge 1$ and draw a ray of $y=2-x$ on a interval $x<-1$.

The graph of $f\left(x\right)$ is shown below in Figure 1.

From Figure 1, it is observed that graph has a parabola with center $\left(0,0\right)$ and passing through the points $\left(-1,1\right)$ and $\left(1,1\right)$. Moreover, the graph has a ray which starts at $\left(1,1\right)$.

Notice that the left-sided limit and the right-sided limit are not same when $a=-1$.

That is, $\underset{x\to -1^{-}}{\lim }f\left(x\right)\ne \underset{x\to -1^{+}}{\lim }f\left(x\right)$.

Therefore, it can be concluded that $\underset{x\to a}{\lim }f\left(x\right)$ exists for every value of a excluding at the point $a=-1$.