Let $f$ be a real valued function and $|f(x)| \le x^2\cdot C + o(x^3)$ as $x\to 0$, where $C \ge 0$ is a constant independent of x. Is it true that there is a $x_0$ such that for all $x\in [0,x_0]$ : $|f(x)| \le x^2\cdot \hat C$ for a constant $\hat C$?
My answer is yes. Since by definition we have for all $g\in o(x^3): \lim_{x\to0}g(x)/x^3 = 0$ thus also $\lim_{x\to0} g(x)/x^2 = 0$ (by product of limits) which means $g\in o(x^2)$. Now $g\in o(x^2)$ implies there is a $x_0 \gt 0$ such that for all $x \le x_0: g(x) \le x^2$ (else the limit $\lim_{x\to0} g(x)/x^2 \ne 0$ or wouldn't exist). Then for all $x\in [0,x_0]: |f(x)| \le x^2\cdot C + o(x^3)\le x^2\cdot C + x^2 \le x^2(1+C) =: x^2 \hat C$. Is my reasoning correct?