Chapter 11.2, Problem 39E

To determine

To find: The value of the $\underset{x\to 1}{\lim }\frac{x-1}{\ln x}$ using table feature of a graphing utility. Also confirm your result graphically.

Answer to Problem 39E

The value of the limits is $\underset{x\to 1}{\lim }\frac{x-1}{\ln x}=1$

Explanation of Solution

Given information:

Given expression of limit as $\underset{x\to 1}{\lim }\frac{x-1}{\ln x}$

Let, $f\left(x\right)=\frac{x-1}{\ln x}$

The value of the function at $x=1.1$ is,

  $\begin{array}{c}f\left(1.1\right)=\frac{1.1-1}{\ln \left(1.1\right)}\\ =\frac{0.1}{\ln \left(1.1\right)}\\ =\frac{0.1}{0.09531}\\ =1.0491\end{array}$

The value of the function at $x=1.01$ is,

  $\begin{array}{c}f\left(1.01\right)=\frac{1.01-1}{\ln \left(1.01\right)}\\ =\frac{0.01}{\ln \left(1.01\right)}\\ =1.005\end{array}$

The value of the function at $x=1.001$ is,

  $\begin{array}{c}f\left(1.001\right)=\frac{1.001-1}{\ln \left(1.1\right)}\\ =\frac{0.001}{\ln \left(1.001\right)}\\ =1.0005\end{array}$

The value of the function at $x=1$ is,

  $\begin{array}{c}f\left(1\right)=\frac{1-1}{\ln \left(1\right)}\\ =\frac{0}{0}\\ =\text{undetermined}\end{array}$

The value of the function at $x=0.9$ is,

  $\begin{array}{c}f\left(0.9\right)=\frac{0.9-1}{\ln \left(0.9\right)}\\ =\frac{-0.1}{\ln \left(0.9\right)}\\ =0.94912\end{array}$

The value of the function at $x=0.99$ is,

  $\begin{array}{c}f\left(0.99\right)=\frac{0.99-1}{\ln \left(0.99\right)}\\ =0.99499\end{array}$

The value of the function at $x=0.999$ is,

  $\begin{array}{c}f\left(0.999\right)=\frac{0.999-1}{\ln \left(0.999\right)}\\ =0.9995\end{array}$

Hence, the complete table is,

  $\begin{array}{cccccccc}x& 1.1& 1.01& 1.001& 1& 0.9& 0.99& 0.999\\ f\left(x\right)& 1.0491& 1.005& 1.0005& \text{undefined}& 0.94912& 0.99499& 0.9995\end{array}$

The value of the limit is, $\underset{x\to 1}{\lim }\frac{x-1}{\ln x}=1$

Graph of the function can be drawn as,

  

From the graph it is clear that $\underset{x\to 1}{\lim }\frac{x-1}{\ln x}=1$