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!

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 $x^5$.  
