<s>[INST] 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.
 [/INST] Recall that by definition
$$f(x)=o(g(x)) \iff f(x)=g(x)\cdot \omega(x) \quad \omega(x) \to 0$$
therefore as $x \to 0$ we have
$$x^2+3x^4+o(x^4)=x^2+3x^4+x^4\cdot \omega(x)=$$$$=x^2+x^3(3x+x\cdot \omega(x)))=x^2+x^3\cdot \omega_1(x)=x^2+o(x^3)$$
but for the limit in your example the key point is that
$$o(x^{n+1})=x\cdot o(x^n) \implies \frac1x \cdot o(x^{n+1})=o(x^n)$$
Therefore we can conclude the limit as as follow
$$\lim_{x\rightarrow0}{\frac{x^2+3x^4+o(x^4)}{x^2+o(x^3)}}=\lim_{x\rightarrow0}{\frac{x^2+o(x^3)}{x^2+o(x^3)}}=\lim_{x\rightarrow0}{\frac{1+o(x)}{1+o(x)}} = \frac{1+0}{1+0}=1$$
or also
$$\lim_{x\rightarrow0}\frac{x^2+3x^4+o(x^4)}{x^2+o(x^3)}=\lim_{x\rightarrow0} \frac{1+3x^2+o(x^2)}{1+o(x)}=\frac{1+0+0}{1+0}= 1$$
</s>