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I know the definition of order estimates,

For the solution to apply we need to show:

$ \displaystyle\lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = A \neq 0\space or \space \infty$

But how can one show this?

JimmyK4542
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jugugularstrik
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  • Do taylor expansion about $\cos(x)$, $i.e$ $ \cos(x) = 1- \frac{x^2}{2} + \cdots$, – phy_math May 20 '15 at 04:53
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    In this case, we can say $cos(x) = 1 + \mathcal{O}(x^2)$

    and then do $\lim_{x \to 0} \frac{1 - 1 + \mathcal{O}(x^2)}{x^2} = 1$ ?

     – jugugularstrik May 20 '15 at 04:56
  • Do you know what $\mathcal{O}(x^2)$ means? – Nikita Evseev May 20 '15 at 05:02
  • My belief is that it is used to describe some equation. $\mathcal{O}(x^2)$ in particular means that the leading terms are $x^2$ when $|x| << 1$ – jugugularstrik May 20 '15 at 05:10
  • Using taylor expansion $\lim_{x\rightarrow 0} \frac{1-\cos(x)}{x^2} = \lim_{x\rightarrow 0} \frac{1-(1-\frac{x^2}{2} + \frac{x^4}{4!} + \cdots)}{x^2} = \lim_{x\rightarrow 0} \frac{1}{2} -\frac{x^2}{4!} + \cdots =\frac{1}{2} + \lim_{x \rightarrow 0} O(x^2) = \frac{1}{2}$ This is the same result for below answer by JimmyK4542 which states $\lim{x \rightarrow 0} \frac{1-\cos(x)}{x^2} = \lim{x\rightarrow 0} \frac{\sin^2(x)}{x^2} \frac{1}{1+\cos(x)} = 1 * \frac{1}{2} =\frac{1}{2}$. – phy_math May 20 '15 at 05:35
  • In other words, the relation of $\frac{\sin{x}}{x}$, $\frac{1-\cos(x)}{x}$ is the typical ways to construct the taylor series of certain function. You can find this materials in any Calculus textbook. – phy_math May 20 '15 at 05:40
  • Thanks Phy math, this is what I needed – jugugularstrik May 20 '15 at 07:05

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Hint: If you already know that $\displaystyle\lim_{x \to 0}\dfrac{\sin x}{x} = 1$, then you can use the following: $\dfrac{1-\cos x}{x^2} = \dfrac{1-\cos x}{x^2} \cdot \dfrac{1+\cos x}{1+\cos x} = \dfrac{1-\cos^2 x}{x^2(1+\cos x)} = \dfrac{\sin^2 x}{x^2(1+\cos x)} = \left[\dfrac{\sin x}{x}\right]^2\dfrac{1}{1+\cos x}$.

JimmyK4542
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