6
$\begingroup$

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 ).

$\endgroup$
  • $\begingroup$ How did you define the cosine? If you defined it by the power series, you can just plug it in. If you didn't, you can still use it, but calculating the power series and using lhospital is basically the same, so I don't know if you want to solve the problem in that way $\endgroup$ – klirk Dec 1 '17 at 17:29
4
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ And how do you calculate hte limit of the sines? $\endgroup$ – klirk Dec 1 '17 at 17:31
  • $\begingroup$ @klirk: $lim_{t→0}\left({{sin(at)}\over{at}}\right) = lim_{t→0}\left({{acos(at)}\over{a}}\right)=cos(0)=1$. $\endgroup$ – Angel Politis Feb 6 '18 at 22:53
  • $\begingroup$ @AngelPolitis But here you used l'Hospital. I asked this to point out, that these limits can be calculated purely geometrical. In fact, this is how one usually proves that $d/dx sin(x)= cos(x)$. So using lhospital for that limit is circular. $\endgroup$ – klirk Feb 7 '18 at 17:49
  • $\begingroup$ Well, I can't post a geometrical calculation here in the comments @klirk 😊 $\endgroup$ – Angel Politis Feb 7 '18 at 18:11
1
$\begingroup$

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}

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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}

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.