Question

use lhôpitals rule to find the limit.

$$lim _ { \theta ightarrow \frac { pi } { 2 } } \frac { 3 - 3 sin \theta } { 7 + 7 cos 2 \theta }$$

$$lim _ { \theta ightarrow \frac { pi } { 2 } } \frac { 3 - 3 sin \theta } { 7 + 7 cos 2 \theta } = square$$

(type an exact answer.)

Explanation:

Step1: Check indeterminate form

Substitute \(\theta=\frac{\pi}{2}\) into \(\frac{3 - 3\sin\theta}{7+7\cos2\theta}\).
\(\sin\frac{\pi}{2}=1\), \(\cos(2\times\frac{\pi}{2})=\cos\pi=- 1\).
\(\frac{3-3\times1}{7 + 7\times(-1)}=\frac{0}{0}\), which is an indeterminate form.

Step2: Apply L'Hôpital's Rule

Differentiate the numerator and denominator.
The derivative of \(y = 3-3\sin\theta\) is \(y^\prime=-3\cos\theta\).
The derivative of \(y = 7 + 7\cos2\theta\) is \(y^\prime=-14\sin2\theta\).
So, \(\lim_{\theta
ightarrow\frac{\pi}{2}}\frac{3 - 3\sin\theta}{7+7\cos2\theta}=\lim_{\theta
ightarrow\frac{\pi}{2}}\frac{-3\cos\theta}{-14\sin2\theta}\)

Step3: Simplify and substitute

Use the double - angle formula \(\sin2\theta = 2\sin\theta\cos\theta\).
\(\lim_{\theta
ightarrow\frac{\pi}{2}}\frac{-3\cos\theta}{-14\sin2\theta}=\lim_{\theta
ightarrow\frac{\pi}{2}}\frac{-3\cos\theta}{-14\times2\sin\theta\cos\theta}\)
Cancel out \(\cos\theta\) (since \(\cos\theta
eq0\) when taking the limit as \(\theta
ightarrow\frac{\pi}{2}\) and \(\theta
eq\frac{\pi}{2}\) in the limit process).
\(\lim_{\theta
ightarrow\frac{\pi}{2}}\frac{3}{28\sin\theta}\)
Substitute \(\theta=\frac{\pi}{2}\), \(\sin\frac{\pi}{2}=1\)

Answer:

\(\frac{3}{28}\)