Question

use lhôpitals rule to find the following limit.

$$\lim_{\theta\to\pi/2}\frac{14\theta - 7\pi}{\cos(2\pi - \theta)}$$

$$\lim_{\theta\to\pi/2}\frac{14\theta - 7\pi}{\cos(2\pi - \theta)}=\square$$ (type an int or a fraction.)

Explanation:

Step1: Check the form of the limit

When \(\theta=\frac{\pi}{2}\), the numerator \(14\theta - 7\pi=14\times\frac{\pi}{2}-7\pi = 7\pi - 7\pi=0\).
The denominator \(\cos(2\pi-\theta)=\cos(2\pi - \frac{\pi}{2})=\cos\frac{3\pi}{2}=0\).
So, it is in the \(\frac{0}{0}\) form, and we can apply L'Hôpital's rule.

Step2: Differentiate the numerator and denominator

Differentiate the numerator: \((14\theta - 7\pi)'=14\).
Differentiate the denominator: \((\cos(2\pi-\theta))'=-\sin(2\pi - \theta)\times(- 1)=\sin(2\pi - \theta)\).
By L'Hôpital's rule, \(\lim_{\theta
ightarrow\frac{\pi}{2}}\frac{14\theta - 7\pi}{\cos(2\pi-\theta)}=\lim_{\theta
ightarrow\frac{\pi}{2}}\frac{14}{\sin(2\pi - \theta)}\).

Step3: Evaluate the new limit

Substitute \(\theta = \frac{\pi}{2}\) into \(\frac{14}{\sin(2\pi - \theta)}\).
\(\sin(2\pi-\theta)=-\sin\theta\), when \(\theta=\frac{\pi}{2}\), \(\sin(2\pi-\frac{\pi}{2})=-\sin\frac{\pi}{2}=- 1\).
So, \(\lim_{\theta
ightarrow\frac{\pi}{2}}\frac{14}{\sin(2\pi - \theta)}=\frac{14}{-1}=-14\).

Answer:

\(-14\)