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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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    They are $o(x^5)$.2012-12-11
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    isn't $x^3o(x^4)=o(x^7)$?2012-12-11
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    Sure is. But to say it is $o(x^5)$ is a *weaker* statement. If $f(x)/x^7$ approaches $0$ as $x\to 0$, then certainly $f(x)/x^5$ approaches $0$. You are accustomed to bigger powers of $x$ being bigger. That's true if $x\gt 1$. But not for $0\lt x\lt 1$.2012-12-11
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    Sorry I am a bit confused. In the following question (http://math.stackexchange.com/questions/250926/little-o-and-its-properties#comment551648_250926) I asked why $x^5 = o(x^2)$ as $x\to0$... and I was told in the comments that $o(f(x))$ means "very smaller than", so as $x\to0$, $x^5$ will definitely be smaller than $x^2$. Can you please elaborate your statement "... is a weaker statement."2012-12-11
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    If $f9x)$ goes to $0$ faster than $x^7$ as $x\to 0$, then $f(x)$ approaches $0$ faster than $x^5$. Write $f(x)/x^5$ as $(x^2)(f(x)/x^7)$. The second part approaches $0$, and the $x^2$ gives it an extra push towards $0$.2012-12-11
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    Thank you, I finally understood. I'll try to write down what I understood: I know that $x^n*x^k = o(x^n)$ for every $k > 0$. So in this case the lowest power is $x^5$. I ignore the others because when the function gets closer to zero ($x\to0$), their "influence" on the value on the function is "masked" by the value of x^5, because it is by far larger than bigger powers of $x$. Am I right?2012-12-11
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    Yes, you have said it. It is the opposite of what happens with large $x$, where big exponents overwhelm small ones.2012-12-11
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    Thank you! How can I mark this post as answered?2012-12-11
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    You can give a brief answer yourself, and accept it. If you don't want to do that, I can write briefly.2012-12-11
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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

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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 $x^5$.