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In class the professor wrote the following limit:

$\lim_{x\to 0} \frac{\sinh^2 (x) -x^2}{x^4}$

So he "expanded" (sorry for my English) the MacLaurin's formula for $\sinh x$ up to the 3rd power, and got: $x + \frac{x^3}{3!} + o(x^4)$

When he squared the MacLaurin's polynomial, he wrote the following steps: $(\sinh x)^2 = (x + \frac{x^3}{3!} + o(x^4))^2 = x^2 + (\frac{x^3}{3!})^2 + (o(x^4))^2 + 2x\frac{x^3}{3!} + 2xo(x^4) + 2\frac{x^3}{3!}o(x^4) = x^2 + \frac{x^4}{3} + o(x^5)$

Now, my question is: why he used $o(x^5)$? As far as I know, when $x\to0$, you have to consider the "$o$" with the highest power, because there is a $o(x^7)$ and an $o(x^8)$

Thank you!

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    It would be better if you write it, and I will mark your answer as accepted. Otherwise you'll lose the possibility to get reputation ;-)2012-12-11

1 Answers 1

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The $o(x^7)$ and $o(x^8)$ are "absorbed" into the $o(x^5)$.

Suppose for example that $f(x)=o(x^7)$. Formally, what that means is that $\lim_{x\to 0} \frac{f(x)}{x^7}=0.$ We show that $f(x)=o(x^5)$. Suppose that $f(x)=o(x^7)$. Then $\lim_{x\to 0}\frac{f(x)}{x^5}=\lim_{x\to 0}x^2\frac{f(x)}{x^7}=(0)(0)=0.$

Informally, if $f(x)$ goes to $0$ faster than $x^7$, then it certainly goes to $0$ faster than $x^5$.