# Exercise about MacLaurin's polynomial and small-o

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 $o(x^5)$. – André Nicolas Dec 11 '12 at 6:37
isn't $x^3o(x^4)=o(x^7)$? – Umar Jamil Dec 11 '12 at 6:38
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 $0\lt x\lt 1$. – André Nicolas Dec 11 '12 at 6:41
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 $0$. – André Nicolas Dec 11 '12 at 6:51
Yes, you have said it. It is the opposite of what happens with large $x$, where big exponents overwhelm small ones. – André Nicolas Dec 11 '12 at 16:25

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 $x^5$.