Order of Infinitesimal of $1-\cos \alpha$ relative to $\alpha$

Question

Exercise

Let the central angle $\alpha$ of a circle of sector $ABO$ with radius $R$ tend to zero. Determine the order of the infinitesimal relative to the infinitesimal $\alpha$ of the line $CD$.

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Attempt

As I was typing up my attempt, I had a revelation, and solved the whole thing! I've decided to put my solution as and answer down below.

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Answer 1

Solution

$DC(\alpha) = R(1 - \cos \alpha)$

I must find $n$ such that $\lim\limits_{\alpha \to 0}{\frac{1-\cos \alpha}{\alpha^n}} \neq 0$.

$\lim\limits_{\alpha \to 0}{\frac{1-\cos \alpha}{\alpha^n}}$ $= \lim\limits_{\alpha \to 0}{\frac{1-\cos \alpha}{\alpha^n}\frac{1+\cos \alpha}{1+\cos \alpha}}$ $= \lim\limits_{\alpha \to 0}{\frac{1-\cos^2 \alpha}{\alpha^n(1+\cos \alpha)}}$ $= \lim\limits_{\alpha \to 0}{\frac{\sin^2 \alpha}{\alpha^n(1+\cos \alpha)}}$ $= \lim\limits_{\alpha \to 0}{\frac{1}{\alpha^{n-2}(1+\cos \alpha)}}$

If $n = 2$, then $\alpha^{n-2}=1$, so $\lim\limits_{\alpha \to 0}{\frac{1}{\alpha^{n-2}(1+\cos \alpha)}}$ $= \lim\limits_{\alpha \to 0}{\frac{1}{1+\cos \alpha}}$ $= \frac{1}{1+\cos 0}$ $ = \frac{1}{2}$

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Answer

$$n = 2$$

The order of the infinitesimal $1-\cos \alpha$ relative to the infinitesimal $\alpha$ is $2$.