# How to compute the limit $\lim\limits_{\alpha \to \pi} \frac{\sin{\alpha}}{\sqrt{2(1+\cos{\alpha})}}$?

My question is simply how do we compute the limit $$\lim\limits_{\alpha \to \pi} \frac{\sin{\alpha}}{\sqrt{2(1+\cos{\alpha})}}$$?

For context on where this came from, consider the following problem.

Spivak, Chapter 11, problem 18:

Ecological Ed must cross a circular lake of radius 1 mile. He can row across at 2mph or walk around at 4mph, or he can row part way and walk the rest. What route should he take so as to

i) see as much scenery as possible

i) seems to be asking what the longest path is. The solution manual says that this is obviously the going around the semicircle, ie walking around. I'd like to prove this.

$$a^2=c^2+d^2$$ $$d=\sqrt{a^2-c^2}$$ $$r^2=c^2+(a+d)^2$$ $$=2a^2+2a\sqrt{a^2-c^2}$$

$$\sin{\alpha}=\frac{c}{a} \implies c =a\sin{\alpha}$$

Therefore

$$r^2(\alpha)=2a^2+2a\sqrt{a^2-a^2\sin^2{\alpha}}$$ $$=2a^2(1+\sqrt{1-\sin^2{\alpha}})$$ $$=2a^2(1+\cos{\alpha})$$

Also

$$w(\alpha)=\alpha a$$

The total distance is therefore

$$l(\alpha)=r(\alpha)+w(\alpha)$$ $$=a\sqrt{2(1+\cos{\alpha}} +a\alpha$$ $$=a(\alpha+\sqrt{2(1+\cos{\alpha}})$$

Taking the derivative

$$l'(\alpha)=a(1+\frac{-2\sin{\alpha}}{2\sqrt{2(1+\cos{\alpha}})}), \alpha \neq \pi$$

$$=a(1-\frac{\sin{\alpha}}{\sqrt{2(1+\cos{\alpha})}}), \alpha \neq \pi$$

$$l'(\alpha)=0 \implies \sin{\alpha}=\sqrt{2(1+\cos{\alpha})}, \alpha \neq \pi$$

If we square both sides

$$\sin^2{\alpha}=2(1+\cos{\alpha}$$ $$1-\cos^2{\alpha}=2(1+\cos{\alpha})$$ $$\cos^2{\alpha}+2\cos{\alpha}+1=0$$

$$\Delta=4-4=0$$ $$\cos{\alpha}=\frac{-2}{2}=-1$$ $$\alpha=\pi$$

Therefore, since $$l'$$ isn't defined at $$\pi$$, this isn't a critical point. In fact we can see that for $$\alpha \in [0,\pi)$$, $$l'>0$$. This would seem to indicate that the longest trajectory is at $$\pi$$, one of the endpoints of the domain of $$l$$.

$$l(\pi)=\pi a = \frac{2\pi a}{2}$$

which makes sense. (note that $$l(0)=2a$$).

I am curious about what happens to $$l'$$ when $$\alpha$$ approaches $$\pi$$. To that end, I need to compute the limit

$$\lim\limits_{\alpha \to \pi} \frac{\sin{\alpha}}{\sqrt{2(1+\cos{\alpha})}}$$

But how? L'Hopital doesn't seem to work.

EDIT: My calculations, based on the hint given by Christophe Leuridan:

$$1+\cos{\alpha}=2\cos^2{\alpha/2}$$ $$\sin{\alpha}=2\sin{\alpha/2}\cos{\alpha/2}$$

$$\lim\limits_{\alpha \to \pi} \frac{\sin{\alpha}}{\sqrt{2(1+\cos{\alpha})}}$$

$$=\lim\limits_{\alpha \to \pi} \frac{2\sin{\alpha/2}\cos{\alpha/2}}{\sqrt{2 \cdot 2\cos^2{\alpha/2}}}$$

$$=\lim\limits_{\alpha \to \pi} \sin{\alpha/2}$$

$$\sin{\pi/2}$$

$$=1$$

EDIT: The above calculations are incorrect, since I incorrectly cancelled $$\cos{\alpha/2}$$ with $$|\cos{\alpha/2}|$$. See comment below for correct version, the result of which is that the limit doesn't exist.

• You wrote $\displaystyle \lim_{x\to \pi}\frac{\sin(\alpha/2)\cos(\alpha/2)}{\sqrt{\cos^{2}(\alpha/2)}}=\lim_{x\to \pi}\sin(\alpha/2)$ why? I think it should be $\displaystyle \lim_{x\to \pi}\frac{\sin(\alpha/2)\cos(\alpha/2)}{|\cos(\alpha/2)|}$. For all $x\in \mathbb{R}$ we have $\sqrt{x^2}=|x|$. In fact, setting $x=\alpha/2$ the limit is the same that $\displaystyle \lim_{x\to \pi/2}\frac{\sin x\cos x}{|\cos x|}$. Commented Apr 23, 2022 at 22:15
• True, I cancelled $\cos{\alpha}$ with $|\cos{\alpha}|$, which is incorrect. Therefore $\lim\limits_{\alpha \to (\frac{\pi}{2})^+} \frac{\sin{\alpha}\cos{\alpha}}{|\cos{\alpha}|}=1$ and $\lim\limits_{\alpha \to (\frac{\pi}{2})^-}\frac{\sin{\alpha}\cos{\alpha}}{|\cos{\alpha}|}=-1$, so $\lim\limits_{\alpha \to (\frac{\pi}{2})} \frac{\sin{\alpha}\cos{\alpha}}{|\cos{\alpha}|}$ doesn't exist.
– xoux
Commented Apr 24, 2022 at 0:01
• Correct: more concisely, the one-sided limits are$$\lim_{\alpha\to\pi^\pm}\frac{\sin\alpha}{\sqrt{2(1+\cos\alpha)}}=\lim_{\alpha\to\pi^\pm}\operatorname{sgn}\left|\cos\frac{\alpha}{2}\right|=\mp1.$$
– J.G.
Commented Apr 24, 2022 at 9:12
• With Taylor series you obtain (as $\alpha\to\pi$): $$\cos(\alpha)=-1+\frac{(\alpha-\pi)^2}2+O((\alpha-\pi)^4)$$ and $$\sin(\alpha) = -(\alpha-\pi)+O((\alpha-\pi)^3)$$ so $$\frac{\sin\alpha}{\sqrt{2(1+\cos(\alpha))}}=\frac{\pi-\alpha+O((\alpha-\pi)^3)}{\lvert{\alpha-\pi}\rvert\sqrt{1+O((\alpha-\pi)^2)}}=\frac{\pi-\alpha}{\lvert\alpha-\pi\rvert (1+O(\alpha-\pi))}+O((\alpha-\pi)^2)\to\mp 1$$ as $\alpha\downarrow\pi$ or $\alpha\uparrow\pi$. Commented Apr 24, 2022 at 9:14

Hint: substitute $$\beta=\alpha-\pi$$.

• The limit becomes $\lim\limits_{\beta \to 0} \frac{\sin{\beta+\pi}}{\sqrt{2(1+\cos{\beta+\pi}}}$, for $beta \in [-\pi,0]$. How does this facilitate the computation?
– xoux
Commented Apr 23, 2022 at 21:06
• @evianpring: Expand $\sin(\beta+\pi)$ and $\cos(\beta+\pi)$ using the trigonometric addition formulae, and the limit becomes $\lim_{\beta\to 0}\frac{-\sin\beta}{\sqrt{2(1-\cos\beta)}}$. Now just substitute the series expansions of $\sin$ and $\cos$. Commented Apr 23, 2022 at 23:21
• Note: you get $\sqrt{\beta^2}$ in the denominator, which I carelessly evaluated as $\beta$. As user1027216's answer points out, it shoud be $|\beta|$, so the expression tends to $+1$ or $-1$ depending on the direction that $\beta$ approaches $0$. Commented Apr 24, 2022 at 8:57

Hint: use the equalities $$\sin\alpha = 2\sin(\alpha/2)\cos(\alpha/2)$$ and $$1+\cos\alpha = 2\cos^2(\alpha/2)$$.

• Using this hint I was able to reach the result that the limit is 0.
– xoux
Commented Apr 23, 2022 at 20:51
• My impression is that the limit as $\alpha \to \pi-$ is 1 whereas the limit as $\alpha \to \pi+$ is -1, since for $\alpha \in (0,2\pi) \setminus \{\pi\}$ the fraction simplifies into $\sin(\alpha/2)\mathrm{sign}(\cos(\alpha/2))$. Commented Apr 23, 2022 at 21:16
• Sorry, yes the limit is 1. The limit that is 0 is the limit $\lim\limits_{\alpha \to \pi} l'(\alpha)=\lim\limits_{\alpha \to \pi} a(1-\frac{\sin{\alpha}}{\sqrt{2(1+\cos{\alpha})}})$, from the Spivak problem.
– xoux
Commented Apr 23, 2022 at 21:30
• However, when I put this limit into Maple, it returns "undefined".
– xoux
Commented Apr 23, 2022 at 21:31

Since

• $$\displaystyle \lim_{\alpha \to \pi^{-}}\frac{1}{\sqrt{2}}\frac{\sin\alpha}{\sqrt{1+\cos \alpha}}=\lim_{x\to \pi^{-}}\frac{1}{\sqrt{2}}\sqrt{\frac{\sin^{2}\alpha\overset{\alpha\to \pi^-}{\to 0}}{1+\cos\alpha\overset{\alpha\to \pi^-}{\to 0}}}=1$$.

• $$\displaystyle \lim_{\alpha \to \pi^{+}}\frac{1}{\sqrt{2}}\frac{\sin\alpha}{\sqrt{1+\cos \alpha}}=-\lim_{x\to \pi^{+}}\frac{1}{\sqrt{2}}\sqrt{\frac{\sin^{2}\alpha\overset{\alpha\to \pi^+}{\to 0}}{1+\cos\alpha\overset{\alpha\to \pi^+}{\to 0}}}=-1$$.

Therefore $$\displaystyle \lim_{x\to \pi} \frac{\sin \alpha}{\sqrt{2(1+\cos\alpha)}}$$ doesn't exist.

• @user1027216 Rephrase your answer the way you did it in the comments. It is much cleaner that way.
– Gary
Commented Apr 24, 2022 at 0:01
• @TonyK For $x$ slightly greater than $\pi$, we have $\displaystyle \frac{\sin x}{\sqrt{1+\cos x}}=-\sqrt{\frac{\sin^2 x}{1+\cos x}}$ isn't correct? I think it's correct but of course if I have made a mistake so we can change my answer. I am here to learn. Commented Apr 24, 2022 at 0:38
• @Gary Of course, thanks for the recommendation, I really appreciate it. At the moment I am wondering if my submitted answer is not correct or doesn't make sense? If it's incorrect I'll gladly rewrite my version of the correction I gave in the comments to the OP. Commented Apr 24, 2022 at 0:41
• @TonyK More formally I'm using that $\sqrt{\sin^{2}x}\underset{\pi^+}{\sim}-\sin x$ and $\sqrt{\sin^{2}x}\underset{\pi^{-}}{\sim}\sin x$. I hope it's clear now. Then inside of the mapping $x\mapsto \sqrt{x}$ I'm using L'Hôpital rule. Commented Apr 24, 2022 at 1:40
• @user1027216: You are correct. I apologise. Commented Apr 24, 2022 at 8:51