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!