Chapter 11.1, Problem 25E

To determine

To find: The value of the $\underset{x\to 2}{\lim }\frac{\ln \left(2x-3\right)}{x-2}$ using table feature of a graphing utility. Also confirm your result graphically.

Answer to Problem 25E

The value of the limits is $\underset{x\to 2}{\lim }\frac{\ln \left(2x-3\right)}{x-2}=2$

Explanation of Solution

Given information:

Given expression of limit as $\underset{x\to 2}{\lim }\frac{\ln \left(2x-3\right)}{x-2}$

Let, $f\left(x\right)=\frac{\ln \left(2x-3\right)}{x-2}$

The value of the function at $x=2.1$ is,

  $\begin{array}{c}f\left(2.1\right)=\frac{\ln \left(2\times 2.1-3\right)}{2.1-2}\\ =\frac{\ln \left(4.2-3\right)}{0.1}\\ =\frac{\ln \left(1.2\right)}{0.1}\\ =1.8232\end{array}$

The value of the function at $x=2.01$ is,

  $\begin{array}{c}f\left(2.01\right)=\frac{\ln \left(2\times 2.01-3\right)}{2.01-2}\\ =\frac{\ln \left(4.02-3\right)}{0.01}\\ =\frac{\ln \left(1.02\right)}{0.01}\\ =1.9803\end{array}$

The value of the function at $x=2.001$ is,

  $\begin{array}{c}f\left(2.001\right)=\frac{\ln \left(2\times 2.001-3\right)}{2.001-2}\\ =\frac{\ln \left(4.002-3\right)}{0.001}\\ =\frac{\ln \left(1.002\right)}{0.001}\\ =1.9980\end{array}$

The value of the function at $x=2$ is,

  $\begin{array}{c}f\left(2\right)=\frac{\ln \left(2\times 2-3\right)}{2-2}\\ =\frac{0}{0}\\ =\text{undefined}\end{array}$

The value of the function at $x=1.9$ is,

  $\begin{array}{c}f\left(1.9\right)=\frac{\ln \left(2\times 1.9-3\right)}{1.9-2}\\ =\frac{\ln \left(3.8-3\right)}{-0.1}\\ =\frac{\ln \left(0.8\right)}{-0.1}\\ =2.2314\end{array}$

The value of the function at $x=1.99$ is,

  $\begin{array}{c}f\left(1.99\right)=\frac{\ln \left(2\times 1.99-3\right)}{1.99-2}\\ =\frac{\ln \left(3.98-3\right)}{-0.01}\\ =\frac{\ln \left(0.98\right)}{-0.01}\\ =2.02027\end{array}$

The value of the function at $x=1.999$ is,

  $\begin{array}{c}f\left(1.999\right)=\frac{\ln \left(2\times 1.999-3\right)}{1.999-2}\\ =\frac{\ln \left(3.998-3\right)}{-0.001}\\ =\frac{\ln \left(0.998\right)}{-0.01}\\ =2.002003\end{array}$

Hence, the complete table is,

  $\begin{array}{cccccccc}x& 2.1& 2.01& 2.001& 2& 1.9& 1.99& 1.999\\ f\left(x\right)& 1.8232& 1.9803& 1.9980& \text{undefined}& 2.2314& 2.02027& 2.002003\end{array}$

The value of the limit is, $\underset{x\to 2}{\lim }\frac{\ln \left(2x-3\right)}{x-2}=2$

Graph of the function can be drawn as,

  

From the graph it is clear that $\underset{x\to 2}{\lim }\frac{\ln \left(2x-3\right)}{x-2}=2$