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

I'm trying to find the limit:

$$\lim_{x \to \frac{\pi}{2}}\left(\frac{\cos(5x)}{\cos(3x)}\right)$$

By L'Hospital's rule it is $-\frac{5}{3}$ but I'm trying to solve it without using L'Hospital rule.

What I tried:

1. Write $\frac{\cos(5x)}{\cos(3x)}$ as $\frac{\cos(4x+x)}{\cos(4x-x)}$ and then using the formula for $\cos(A+B)$.




2. Write $\cos(x)$ as $\sin\left(x - \frac{\pi}{2}\right)$.

But I didn't have success with those methods (e.g. in the first one I got the same expression $\frac{\cos(5x)}{\cos(3x)}$ again ).

Answer

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