Is $1-\cos^2(x)$ a greater infinitesimal than $\sin^3(x)$ as $\to 0+$?

Question

What I did was to see that both are $\sin$ functions and the first one acts as $x^2$ while the second one goes to $x^3$. So the order of infinitesimal of the first one is $2$ and for the second it is $3$ which means that the second one is an infinitesimal greater than the first one. But our exam portal says the opposite. I just want to make sure I am not confusing the terminology.

Answer 1

The confusion here comes from unclear notation in the question. The author means $f(x) = 1 - \cos(x^2)$ and not $f(x) = 1-\cos(x)^2$ as you have assumed. You can see this from the author using the notation $\sin^3(x)$ to denote $\sin(x)^3$.

Using this definition of $f$ gives $f(x) \sim \frac{x^4}{2}$ which goes to zero faster than $g(x) = \sin(x)^3$.

Answer 2

Assuming $f=1-cos^2x$ you were right, the infinitesimal order of $f$ should be 2 and the infinitesimal order of $g$ is 3, then the option given should be wrong.

EDIT

As noted, $f=1-cos(x^2)=\frac{x^4}2+o(x^4)$ then the infinitesimal order of $f$ is 4.

Answer 3

The question is about $1-cos(x^2)$ not $1-cos^2(x).$

Note that $1-cos(x^2)\sim x^4/2$ which has a greater order of infinitesimal than $sin^3(x)$