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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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.