evaluate the limit using lhopitals rule lim x→π/2 8 cos(−3x) sec(9x)

Explanation:

Step1: Recall secant - cosine relation

We know that $\sec(9x)=\frac{1}{\cos(9x)}$, so the function becomes $y = 8\cos(-3x)\frac{1}{\cos(9x)}=\frac{8\cos(-3x)}{\cos(9x)}$. When $x\rightarrow\frac{\pi}{2}$, we have $\cos(-3x)=\cos(-\frac{3\pi}{2}) = 0$ and $\cos(9x)=\cos(\frac{9\pi}{2})=0$, which is in the $\frac{0}{0}$ indeterminate form.

Step2: Apply L'Hopital's rule

Differentiate the numerator and denominator. The derivative of $u = 8\cos(-3x)$ using the chain - rule: $u^\prime=8(-\sin(-3x))(-3)=24\sin(-3x)$. The derivative of $v=\cos(9x)$ using the chain - rule: $v^\prime=-\sin(9x)\times9=-9\sin(9x)$. So the limit becomes $\lim_{x\rightarrow\frac{\pi}{2}}\frac{24\sin(-3x)}{-9\sin(9x)}$.

Step3: Evaluate the new limit

When $x = \frac{\pi}{2}$, $\sin(-3x)=\sin(-\frac{3\pi}{2}) = 1$ and $\sin(9x)=\sin(\frac{9\pi}{2}) = 1$. Then $\lim_{x\rightarrow\frac{\pi}{2}}\frac{24\sin(-3x)}{-9\sin(9x)}=\frac{24\times1}{-9\times1}=-\frac{8}{3}$.

Answer:

$-\frac{8}{3}$