Chapter 4.5, Problem 21E

To determine

The value of given limit.

Answer to Problem 21E

  $-\frac{1}{{\pi }^2}$ .

Explanation of Solution

Given information:

A limit function is given as,

  $\underset{x\to 1}{\lim }\frac{1-x+\ln x}{1+\cos (\pi x)}$ .

Concept used:

L'Hospital's Rule: If $f$ and $g$ are differentiable in an open interval containing $a$ and $f(a)=g(a)=0$ . Also, $g\text{'}(x)\ne 0$ (except possibly at $a$ ). Then-

  $\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f\text{'}(x)}{g\text{'}(x)}$

This rule is also applicable when $f(a)=\pm \infty \text{ and }g(a)=\pm \infty$ .

Calculation:

Given limit function is, $\underset{x\to 1}{\lim }\frac{1-x+\ln x}{1+\cos (\pi x)}$ .

Let us suppose, $f(x)=1-x+\ln x$ and $g(x)=1+\cos (\pi x)$ .

Here limit converts into an indeterminate form of $\frac{0}{0}$ .

L'Hospital's Rule is applicable here.

Using L'Hospital's Rule,

  $\begin{array}{l}\text{Now, }\underset{x\to 1}{\lim }\frac{1-x+\ln x}{1+\cos (\pi x)}\\ =\underset{x\to 1}{\lim }\frac{f\text{'}(x)}{g\text{'}(x)}\\ =\underset{x\to 1}{\lim }\frac{0-1+\frac{1}{x}}{0-\pi \sin (\pi x)}\\ \text{This again converts into indeterminate form of }\frac{0}{0}.\\ \text{Applying L'Hospital's rule again,}\\ =\underset{x\to 1}{\lim }\frac{-\frac{1}{x^2}}{-{\pi }^2\cos (\pi x)}\\ \text{Plug in the value }x=1\\ =\frac{-\frac{1}{1^2}}{-{\pi }^2\cos (\pi \cdot 1)}\\ =-\frac{1}{{\pi }^2}\end{array}$

Hence, value of given limit function is $-\frac{1}{{\pi }^2}$ .