Chapter 1.1, Problem 16E

(a)

To determine

To complete: the given table for $x\to 0^{-}$ .

Answer to Problem 16E

  $\begin{array}{ccccc}x& -0.1& -0.01& -0.001& -0.0001\\ f\left(x\right)& -1.1& -1.01& -1.001& -1.0001\end{array}$

And

  $\underset{x\to 0^{-}}{\lim }f\left(x\right)=-1$

Explanation of Solution

Given information:

The given table is

The given function is $f\left(x\right)=\frac{x^2-x}{x}$

The given table for $x\to 0^{-}$ is

  $\begin{array}{ccccc}x& -0.1& -0.01& -0.001& -0.0001\\ f\left(x\right)& & & & \end{array}$

Calculation:

Now, evaluate $f\left(x\right)$ for $x=-0.1$ .

  $\begin{array}{c}f\left(-0.1\right)=\frac{{\left(-0.1\right)}^2-\left(-0.1\right)}{\left(-0.1\right)}\\ =-1.1\end{array}$

Evaluate $f\left(x\right)$ for $x=-0.01$ .

  $\begin{array}{c}f\left(-0.01\right)=\frac{{\left(-0.01\right)}^2-\left(-0.01\right)}{\left(-0.01\right)}\\ =-1.01\end{array}$

Evaluate $f\left(x\right)$ for $x=-0.001$ .

  $\begin{array}{c}f\left(-0.001\right)=\frac{{\left(-0.001\right)}^2-\left(-0.001\right)}{\left(-0.001\right)}\\ =-1.001\end{array}$

Evaluate $f\left(x\right)$ for $x=-0.0001$ .

  $\begin{array}{c}f\left(-0.001\right)=\frac{{\left(-0.0001\right)}^2-\left(-0.0001\right)}{\left(-0.0001\right)}\\ =-1.0001\end{array}$

The completed table for $x\to 0^{-}$ is as follows.

  $\begin{array}{ccccc}x& -0.1& -0.01& -0.001& -0.0001\\ f\left(x\right)& -1.1& -1.01& -1.001& -1.0001\end{array}$

Therefore, for $x\to 0^{-}$ the function $f\left(x\right)$ approaches to $-1$ .

Hence, $\underset{x\to 0^{-}}{\lim }f\left(x\right)=-1$

(b)

To determine

To complete: the given table for $x\to 0^{+}$ .

Answer to Problem 16E

  $\begin{array}{ccccc}x& 0.1& 0.01& 0.001& 0.0001\\ f\left(x\right)& -0.9& -0.99& -0.999& -0.9999\end{array}$

And

  $\underset{x\to 0^{+}}{\lim }f\left(x\right)=-1$

Explanation of Solution

Given information:

The given table is

The given function is $f\left(x\right)=\frac{x^2-x}{x}$

The given table for $x\to 0^{+}$ is

  $\begin{array}{ccccc}x& 0.1& 0.01& 0.001& 0.0001\\ f\left(x\right)& & & & \end{array}$

Calculation:

Now, evaluate $f\left(x\right)$ for $x=0.1$ .

  $\begin{array}{c}f\left(0.1\right)=\frac{{\left(0.1\right)}^2-\left(0.1\right)}{\left(0.1\right)}\\ =-0.9\end{array}$

Evaluate $f\left(x\right)$ for $x=0.01$ .

  $\begin{array}{c}f\left(0.01\right)=\frac{{\left(0.01\right)}^2-\left(0.01\right)}{\left(0.01\right)}\\ =-0.99\end{array}$

Evaluate $f\left(x\right)$ for $x=0.001$ .

  $\begin{array}{c}f\left(0.001\right)=\frac{{\left(0.001\right)}^2-\left(0.001\right)}{\left(0.001\right)}\\ =-0.999\end{array}$

Evaluate $f\left(x\right)$ for $x=0.0001$ .

  $\begin{array}{c}f\left(0.001\right)=\frac{{\left(0.0001\right)}^2-\left(0.0001\right)}{\left(0.0001\right)}\\ =-0.9999\end{array}$

The completed table for $x\to 0^{+}$ is as follows.

  $\begin{array}{ccccc}x& 0.1& 0.01& 0.001& 0.0001\\ f\left(x\right)& -0.9& -0.99& -0.999& -0.9999\end{array}$

Therefore, for $x\to 0^{+}$ the function $f\left(x\right)$ approaches to $-1$ .

Hence, $\underset{x\to 0^{+}}{\lim }f\left(x\right)=-1$

From part (a) and part (b) it can be concluded that

  $\underset{x\to 0}{\lim }f\left(x\right)=-1$