Chapter 13.1, Problem 15E

To determine

To estimate: The value of the limit.

Answer to Problem 15E

The value of the limit is $0\cdot 5$

Explanation of Solution

Given information:

Limit is $\underset{x\to 1}{\lim }\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)$

Concept used:

Limit is in $\frac{0}{0}$ form it is called the determinate form then we can use L’ Hospital’s Rule.

Formula used:

  $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}\text{ when }\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{0}{0}\text{ (or) }\frac{\pm \infty }{\pm \infty }$

Calculation:

Limit is $\underset{x\to 0}{\lim }\frac{1}{\ln x}-\frac{1}{x-1}$

  $\underset{x\to 0}{\lim }\frac{x-1-\ln x}{\ln \left(x-1\right)}=\frac{-1-\ln 1}{\ln 1\left(1-1\right)}=\frac{0}{0}\text{ form}$

Therefore,

  $f\left(x\right)=x-1-\ln xg\left(x\right)=\ln x\left(x-1\right)$

  $\therefore$ Differentiation of $\ln x$ is $\frac{1}{x}$

Differentiation of $a.b$ is $\frac{da}{dx}b+\frac{db}{dx}a$

  $\begin{array}{l}f\text{'}\left(x\right)=1-\frac{1}{x}g\text{'}\left(x\right)=\frac{d\left(\ln x\right)}{dx}\cdot \left(x-1\right)+\frac{d\left(x-1\right)}{dx}\cdot \ln x\\ g\text{'}\left(x\right)=\frac{x-1}{x}+\ln x\\ \underset{x\to 1}{\lim }\frac{x-1-\ln x}{x-1+x\ln x}=\frac{1-\frac{1}{x}}{\frac{x-1}{x}+\ln x}=\frac{0}{0}\text{ form}\\ \text{p}\left(x\right)=x-1q\left(x\right)=x-1+x\ln x\\ \text{p'}\left(x\right)=1q\text{'}\left(x\right)=1+\ln x+x\cdot \frac{1}{x}=2+\ln x\\ \underset{x\to 1}{\lim }\frac{x-1-\ln x}{x-1+x\ln x}=\underset{x\to 1}{\lim }\frac{1}{2+\ln x}=\frac{1}{2}\end{array}$

Verification:

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $\underset{x\to 1}{\lim }\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)$ which is required to graph.

Step 4: Press GRAPH button to graph the function and adjust the windows according to the graph.

The graph is obtained as:

  

From graph, it can be observed that points near by $1$ y - coordinate is $0\cdot 5$ nearly.