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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 ).

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  • $\begingroup$ 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, 2017 at 17:29

6 Answers 6

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

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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}

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  • $\begingroup$ And how do you calculate hte limit of the sines? $\endgroup$
    – klirk
    Dec 1, 2017 at 17:31
  • $\begingroup$ @klirk: $lim_{t→0}\left({{sin(at)}\over{at}}\right) = lim_{t→0}\left({{acos(at)}\over{a}}\right)=cos(0)=1$. $\endgroup$ Feb 6, 2018 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, 2018 at 17:49
  • $\begingroup$ Well, I can't post a geometrical calculation here in the comments @klirk 😊 $\endgroup$ Feb 7, 2018 at 18:11
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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}

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

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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}

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

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