use lhôpitals rule to find the following limit. lim θ→π/2 4θ - 2π / cos(2π - θ) lim θ→π/2 4θ - 2π / cos(2π…

Explanation:

Step1: Check indeterminate form

First, substitute $\theta=\frac{\pi}{2}$ into $\frac{4\theta - 2\pi}{\cos(2\pi-\theta)}$. We get $\frac{4\times\frac{\pi}{2}-2\pi}{\cos(2\pi - \frac{\pi}{2})}=\frac{2\pi - 2\pi}{\cos(\frac{3\pi}{2})}=\frac{0}{0}$, which is an indeterminate - form. So, we can apply L'Hopital's rule.

Step2: Differentiate numerator and denominator

Differentiate the numerator $y_1 = 4\theta-2\pi$ with respect to $\theta$. Using the power - rule, $y_1^\prime=\frac{d}{d\theta}(4\theta - 2\pi)=4$. Differentiate the denominator $y_2=\cos(2\pi-\theta)$ with respect to $\theta$. Using the chain - rule, if $u = 2\pi-\theta$, then $y_2=\cos(u)$ and $\frac{dy_2}{d\theta}=\frac{dy_2}{du}\cdot\frac{du}{d\theta}$. We know that $\frac{d}{du}(\cos(u))=-\sin(u)$ and $\frac{du}{d\theta}=-1$. So, $\frac{d}{d\theta}(\cos(2\pi - \theta))=-\sin(2\pi-\theta)\times(-1)=\sin(2\pi-\theta)$.

Step3: Find the new limit

Now, find the limit of the quotient of the derivatives: $\lim_{\theta\rightarrow\frac{\pi}{2}}\frac{4}{\sin(2\pi-\theta)}$. Substitute $\theta = \frac{\pi}{2}$ into $\frac{4}{\sin(2\pi-\theta)}$, we have $\sin(2\pi-\frac{\pi}{2})=\sin(\frac{3\pi}{2})=-1$. So, $\lim_{\theta\rightarrow\frac{\pi}{2}}\frac{4}{\sin(2\pi-\theta)}=-4$.

Answer:

$-4$