Does $x^2+3x^4+o(x^4)=x^2+o(x^3)$ hold true? My textbook asks if the following holds true or false for $x\rightarrow0$:
$$x^2+3x^4+o(x^4)=x^2+o(x^3)$$
I think I understand what the little-o notation means. For $x\rightarrow x_o$ we have:
$$\lim_{x\rightarrow x_0}{\frac{f(x)}{g(x)}=0}$$
and we say $f(x)=o(g(x))$.
So to find an answer to the problem I evaluated the following limit:
$$\lim_{x\rightarrow0}{\frac{x^2+3x^4+o(x^4)}{x^2+o(x^3)}}$$
For $x\rightarrow0$, I think we have
$$\frac{x^2+3x^4+o(x^4)}{x^2+o(x^3)}\sim \frac{x^2+o(x^4)}{x^2+o(x^3)}=\dots$$
Any hints on how to continue? My textbook says the answer should be true. Sorry I couldn't find the appropriate tag for this little-o notation question.

Note that:
$$\lim_{x\rightarrow0}{\frac{x^2+o(x^4)}{x^2+o(x^3)}}=\lim_{x\rightarrow0}{\frac{x^2[1+o(x^2)]}{x^2[1+o(x)]}}=\lim_{x\rightarrow0}{\frac{1+o(x^2)}{1+o(x)}}=\frac{1+0}{1+0}=1.$$
