Find the limit $\lim_{x \to \frac{\pi}{2}}(\frac{\cos(5x)}{\cos(3x)})$ without using L'Hospital's rule

$$\cos(5x)=\sin \left(\frac52 \pi-5x\right)=\sin5\left(\frac{\pi}{2}-x\right)$$ And $$\cos(3x)=-\sin\left(\frac32 \pi-3x\right)=-\sin3\left(\frac{\pi}{2}-x\right)$$

So we set $\frac{\pi}{2}-x=w$

as $x\to \frac{\pi}{2} $ we have $x\to 0$

The given limit can be written as

$$\lim_{w\to 0}\frac{\sin 5w}{-\sin 3w}=-\frac{5}{3}\lim_{w\to 0}\frac{3w\sin 5w}{5w\sin 3w}=-\frac{5}{3}\lim_{w\to 0}\left(\frac{\sin 5w}{5w}\cdot \frac{3w}{\sin3w}\right)=-\frac{5}{3}$$ Hope this can be useful

Write $t=x-\pi/2$, then

\begin{eqnarray}\lim_{x \to \frac{\pi}{2}}\frac{\cos(5x)}{\cos(3x)}&=&\lim_{t \to 0}\frac{\sin(-5t-2\pi)}{\sin(-3t-\pi)} \\&=&\lim_{t \to 0}\frac{-\sin(5t)}{\sin(3t)} \\ &=& \cdot\lim_{t \to 0}\frac{-\sin(5t)}{5t}\cdot \frac{3t}{\sin(3t)}\cdot {5\over 3}\\ &=& -{5\over 3}\cdot\lim_{t \to 0}\frac{\sin(5t)}{5t}\cdot\lim_{t \to 0}\frac{3t}{\sin(3t)}\\ &=& -{5\over 3}\cdot\underbrace{\lim_{t \to 0}\frac{\sin(5t)}{5t}}_{=1}\cdot \Big( \underbrace{\lim_{t \to 0}\frac{\sin (3t)}{3t}}_{=1}\Big)^{-1} \\&=&-{5\over 3}\end{eqnarray}