evaluate the limit using lhopitals rule lim x...

# Explanation: ## Step1: Recall secant - cosine relation $\sec(5x)=\frac{1}{\cos(5x)}$, so the function is $y = 6\frac{\cos(-7x)}{\cos(5x)}$. When $x=\frac{\pi}{2}$, $\cos(-7x)=\cos(-\frac{7\pi}{2})=\cos(\frac{7\pi}{2}) = 0$ and $\cos(5x)=\cos(\frac{5\pi}{2}) = 0$. It is in $\frac{0}{0}$ form, so we can apply L'Hopital's rule. ## Step2: Differentiate numerator and denominator The derivative of $\cos(-7x)$ using the chain - rule: If $u=-7x$, then $\frac{d}{dx}\cos(-7x)=-\sin(-7x)\times(-7) = 7\sin(-7x)$. The derivative of $\cos(5x)$ using the chain - rule: If $u = 5x$, then $\frac{d}{dx}\cos(5x)=-\sin(5x)\times5=-5\sin(5x)$. The derivative of the function $y$ is $y'=6\frac{7\sin(-7x)}{-5\sin(5x)}$. ## Step3: Evaluate the limit of the derivative $\lim_{x\rightarrow\frac{\pi}{2}}6\frac{7\sin(-7x)}{-5\sin(5x)}$. When $x = \frac{\pi}{2}$, $\sin(-7x)=\sin(-\frac{7\pi}{2})=\sin(\frac{\pi}{2}) = 1$ and $\sin(5x)=\sin(\frac{5\pi}{2}) = 1$. Then $\lim_{x\rightarrow\frac{\pi}{2}}6\frac{7\sin(-7x)}{-5\sin(5x)}=6\times\frac{7\times1}{-5\times1}=-\frac{42}{5}$. # Answer: $-\frac{42}{5}$