Chapter 9.2, Problem 6E

To determine

To Find: The value of a limit using L'Hôpital's Rule

Answer to Problem 6E

The value of the limit is $\frac{1}{4}$

Explanation of Solution

Given Information: The limit to be calculated is $\underset{\theta \to \frac{\pi }{2}}{\lim }\frac{1-\sin \theta }{1+\cos 2\theta }$

Concept Used:

  $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(a)=g(a)=0,\text{ then }\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation: At $\theta =\frac{\pi }{2}$ , the value of the expression is $\frac{1-\sin \frac{\pi }{2}}{1+\cos \frac{2\pi }{2}}=\frac{1-1}{1-1}=\frac{0}{0}$

So, apply L'Hôpital's Rule to write

  $\begin{array}{l}\underset{\theta \to \frac{\pi }{2}}{\lim }\frac{1-\sin \theta }{1+\cos 2\theta }=\underset{\theta \to \frac{\pi }{2}}{\lim }\frac{0-cos\theta }{0-2sin2\theta }\\ =\underset{\theta \to \frac{\pi }{2}}{\lim }\frac{cos\theta }{2sin2\theta }\end{array}$

Again, when $\theta =\frac{\pi }{2}$ , the value of $\frac{cos\theta }{2sin2\theta }$ is $\frac{0}{2\times 0}=\frac{0}{0}$

Therefore, apply L'Hôpital's Rule again to write

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

Therefore, the limit is $\frac{1}{4}$