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:


*

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

*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 ). 
 A: $$\cos(5x)=\sin \left(\frac52 \pi-5x\right)=\sin5\left(\frac{\pi}{2}-x\right)$$
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
A: 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}
A: Note that\begin{align}\cos(5x)&=\cos\left(5x-\frac{5\pi}2+\frac{5\pi}2\right)\\&=-\sin\left(5\left(x-\frac\pi2\right)\right)\\&=-5\left(x-\frac\pi2\right)+o\left(\left(x-\frac\pi2\right)^2\right)\end{align}and that, for a similar reason,$$\cos(3x)=3\left(x-\frac\pi2\right)+o\left(\left(x-\frac\pi2\right)^2\right).$$Therefore\begin{align}\lim_{x\to\frac\pi2}\frac{\cos(5x)}{\cos(3x)}&=\lim_{x\to\frac\pi2}\frac{-5\left(x-\frac\pi2\right)+o\left(\left(x-\frac\pi2\right)^2\right)}{3\left(x-\frac\pi2\right)+o\left(\left(x-\frac\pi2\right)^2\right)}\\&=\lim_{x\to\frac\pi2}\frac{-5+o\left(x-\frac\pi2\right)}{3+o\left(x-\frac\pi2\right)}\\&=-\frac53.\end{align}
A: Same method by substitution, but a little shorter with equivalents:
Set $x=\frac\pi 2-u,\;(u\to 0)$. Remember $\sin t\sim_0 t$. We have: 
$$\frac{\cos 5x}{\cos 3x}=\frac{\cos\Bigl(\dfrac{5\pi}2-5u\Bigr)}{\cos\Bigl(\dfrac{3\pi}2-3u\Bigr)}=\frac{\cos\Bigl(\dfrac{\pi}2-5u\Bigr)}{\cos\Bigl(-\dfrac{\pi}2-3u\Bigr)}=\frac{\sin 5u}{-\sin3u}\sim_0\frac{5\not u}{-3\not u}=-\frac53. $$
A: Since
$$
\sin(a-b)=\sin(a)\cdot\cos(b)-\cos(a)\cdot\sin(b)
$$ 
we have 
$$
\sin(a-5\cdot\pi/2)
=
\sin(a)\cdot\cos(5\cdot\pi/2)-\cos(a)\cdot\sin(5\cdot\pi/2)
=
\sin(a)\cdot0-\cos(a)\cdot 1
=
-\cos (a)
$$
and
$$
\sin(a-3\cdot\pi/2)
=
\sin(a)\cdot\cos(3\cdot\pi/2)-\cos(a)\cdot\sin(3\cdot\pi/2)
=
\sin(a)\cdot0-\cos(a)\cdot (-1)
=
+\cos (a)
$$
Now the limit.
\begin{align}
\lim_{x\to \pi/2}\frac{\cos(5x)}{\cos(3x)}
=&
\lim_{x\to \pi/2}\frac{-\sin(5x-5\pi/2)}{\sin(3x-3\pi/2)}
\\\\
=&
-\lim_{x\to \pi/2}\frac{\sin(5(x-\pi/2)}{\sin(3(x-\pi/2)}
\\
\\
=&
-\lim_{t\to 0}\frac{\sin(5t)}{\sin(3t)}
\\
\\
=&
-\lim_{t\to 0}\dfrac{(5t)\cdot \dfrac{\sin(5t)}{(5t)}}{(3t)\dfrac{\sin(3t)}{(3t)}}
\\\\
=&
-5/3\lim_{t\to 0}\dfrac{ \dfrac{\sin(5t)}{(5t)}}{\dfrac{\sin(3t)}{(3t)}}
\\
=&
-5/3
\end{align}
A: hint
$$\cos ((2p+1)x)=$$
$$(-1)^p\sin ((2p+1)(\frac {\pi}{2}-x)) $$
$$\sim (-1)^p (2p+1)(\frac \pi 2-x) \;\;(x\to \pi/2)$$
the limit is $$-\frac {2.2+1}{2.1+1}=-5/3$$
