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I'm asked to find the optimal $k$ such that $\cos{x}-1+x^2/2=O(x^k)\quad (x\to 0)$.

By Taylor's theorem I know that $\cos{x}-1+x^2/2=(1/4!)\xi^4$ for some $\xi\in (0,x)$. So $0<\cos{x}-1+x^2/2<(1/4!)x^4$, that is $\cos{x}-1+x^2/2=O(x^4)$.

My reasoning is as follows: if $0

Next I try to see whether $\cos{x}-1+x^2/2=O(x^{4+\delta})$:

$$\frac{\cos{x}-1+x^2/2}{x^{4+\delta}}=\frac{1}{x^\delta}\frac{\cos{x}-1+x^2/2}{x^4}.$$ But by repeated application of L'Hôpital's, $$\lim_{x\to 0}\frac{\cos{x}-1+x^2/2}{x^4}=1/24,$$ so $\displaystyle\lim_{x\to 0} \frac{\cos{x}-1+x^2/2}{x^{4+\delta}}=\infty$.

This seems to show that $x^4$ is indeed the optimal order.

So is all that correct? Also, $1/24=1/4!$, so surely that's not a coincidence, but I don't see how to arrive at that limit in a different way (I don't know how that $\xi$ above depends on $x$).

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    Isn't $\cos(x)-1+\frac{x^2}{2} < x^2$ for $x$ large? Note that $-2\le \cos(x)-1 \le 0$.2011-03-12
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    @Henry I'm interested in $x\to 0$.2011-03-12
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    Oh okay, yes, that makes sense. I missed that.2011-03-12
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    @Henry B.: your relation is correct for $x\lt \lt 1$ which is what was asked. Large, too, as $\cos(x)-1\le 0$, so $\cos(x)-1+\frac{x^2}{2} \le x^2$ (and equality only obtains for $x=0$)2011-03-12

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Yes, that is all correct.

You're correct that $24=4!$ is not a coincidence here. From Taylor's theorem you can derive a power series expansion for cosine, $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots,$$ so that $$\cos(x)-1+\frac{x^2}{2}=\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots,$$ and $$\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}=\frac{1}{4!}-\frac{x^2}{6!}+\frac{x^4}{8!}-\cdots.$$ The series on the right-hand side of the the last equation is defined and continuous everywhere, and goes to $\frac{1}{4!}$ at $0$.