Chapter 8.2, Problem 6E

To determine

To find: The limit using stronger form of l’Hospital rule.

Answer to Problem 6E

  $\underset{\theta \to \frac{\pi }{2}}{\lim }\frac{1-\sin \theta }{1+\cos \left(2\theta \right)}=\frac{1}{4}$

Explanation of Solution

Given information :

The general form of sequence is as follow:

  $\underset{\theta \to \frac{\pi }{2}}{\lim }\frac{1-\sin \theta }{1+\cos \left(2\theta \right)}$

Concept used:

Suppose,

  $\begin{array}{c}f\left(a\right)=g\left(a\right)\\ =0\end{array}$

Where $f\text{and}g$ are differentiable on an open interval $I$ containing $a$ , and $g\text{'}\left(x\right)\ne 0$ 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 :

Here, $f\left(\theta \right)=1-\sin \theta \text{and}g\left(\theta \right)=1+\cos \left(2\theta \right)$

By putting $\theta =\frac{\pi }{2}$ :

  $f\left(0\right)=0\text{and}g\left(0\right)=0$

This gives indeterminant form $\frac{0}{0}$ .

Applying l’Hospital rule:

  $\begin{array}{c}\underset{\theta \to \frac{\pi }{2}}{\lim }\frac{1-\sin \theta }{1+\cos \left(2\theta \right)}=\underset{\theta \to \frac{\pi }{2}}{\lim }\frac{\frac{d}{d\theta }\left[1-\sin \theta \right]}{\frac{d}{d\theta }\left[1+\cos \left(2\theta \right)\right]}\\ =\underset{\theta \to \frac{\pi }{2}}{\lim }\frac{\left[-\cos \theta \right]}{\left[-2\sin \left(2\theta \right)\right]}\\ =\frac{0}{0}\end{array}$

Differentiating again:

  $\begin{array}{c}\underset{\theta \to \frac{\pi }{2}}{\lim }\frac{\left[-\cos \theta \right]}{\left[-2\sin \left(2\theta \right)\right]}=\underset{\theta \to \frac{\pi }{2}}{\lim }\frac{\frac{d}{d\theta }\left[\cos \theta \right]}{\frac{d}{d\theta }\left[2\sin \left(2\theta \right)\right]}\\ =\underset{\theta \to \frac{\pi }{2}}{\lim }\frac{\left[-\sin \theta \right]}{\left[4\cos \left(2\theta \right)\right]}\\ =\frac{-1}{-4}\\ =\frac{1}{4}\end{array}$

Conclusion:

  $\underset{\theta \to \frac{\pi }{2}}{\lim }\frac{1-\sin \theta }{1+\cos \left(2\theta \right)}=\frac{1}{4}$