Chapter 8, Problem 11RE

To determine

To find: The value of the given limit.

Answer to Problem 11RE

The value of the given limit is $\underset{x\to 0}{\lim }\frac{x\sin x}{\left(1-\cos x\right)}=2$ .

Explanation of Solution

Given information: The limit is $\underset{x\to 0}{\lim }\frac{x\sin x}{\left(1-\cos x\right)}$ .

Theorem used:(L’Hospital’s Rule).

“Suppose that $\underset{x\to a}{\lim }f\left(x\right)=0\text{ and }\underset{x\to a}{\lim }g\left(x\right)=0$ , that f and g are differentiable on an open interval I containing a , and that $g\text{'}\left(x\right)\ne 0$ on I if $x\ne a$ .Then $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$ , if the latter limit exists.”

Calculation:

Apply the limit as t tends to zero as follows.

  $\begin{array}{c}\underset{x\to 0}{\lim }\frac{x\sin x}{\left(1-\cos x\right)}=\frac{0}{1-1}\\ =\frac{0}{0}\\ =\text{Indeterminate form}\end{array}$

Apply the L’Hospital’s rule as follows.

  $\begin{array}{c}\underset{t\to 0}{\lim }\frac{x\sin x}{\left(1-\cos x\right)}=\underset{t\to 0}{\lim }\left(\frac{\frac{d}{dx}\left(x\sin x\right)}{\frac{d}{dx}\left(1-\cos x\right)}\right)\\ =\underset{t\to 0}{\lim }\left(\frac{\sin x\frac{dx}{dx}+x\frac{d}{dx}\left(\sin x\right)}{\frac{d}{dx}\left(1-\cos x\right)}\right)\\ =\underset{t\to 0}{\lim }\left(\frac{\sin x+x\cos x}{\sin x}\right)\end{array}$

On further simplifications.

  $\begin{array}{c}\underset{t\to 0}{\lim }\frac{x\sin x}{\left(1-\cos x\right)}=\frac{0+0}{0}\\ =\frac{0}{0}\\ =\text{Indeterminate form}\end{array}$

Again, apply the L’Hospital’s rule as follows.

  $\begin{array}{c}\underset{t\to 0}{\lim }\frac{x\sin x}{\left(1-\cos x\right)}=\underset{t\to 0}{\lim }\left(\frac{\frac{d}{dx}\left(\sin x+x\cos x\right)}{\frac{d}{dx}\left(\sin x\right)}\right)\\ =\underset{t\to 0}{\lim }\left(\frac{2\cos x-x\sin x}{\cos x}\right)\\ =\frac{2\left(1\right)-0\left(0\right)}{1}\\ =\frac{2-0}{1}\\ =2\end{array}$

Therefore, the value of the given limit is $\underset{x\to 0}{\lim }\frac{x\sin x}{\left(1-\cos x\right)}=2$ .