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

Question

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 1

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

Answer 2

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}

Answer 3

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 $$-\frac {2.2+1}{2.1+1}=-5/3$$

Answer 4

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}

Answer 5

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: $$\frac{\cos 5x}{\cos 3x}=\frac{\cos\Bigl(\dfrac{5\pi}2-5u\Bigr)}{\cos\Bigl(\dfrac{3\pi}2-3u\Bigr)}=\frac{\cos\Bigl(\dfrac{\pi}2-5u\Bigr)}{\cos\Bigl(-\dfrac{\pi}2-3u\Bigr)}=\frac{\sin 5u}{-\sin3u}\sim_0\frac{5\not u}{-3\not u}=-\frac53. $$

Answer 6

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}