Does $x^2+3x^4+o(x^4)=x^2+o(x^3)$ hold true?

Question

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.

Answer 1

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

$$\lim_{x\rightarrow0}\frac{x^2+3x^4+o(x^4)}{x^2+o(x^3)}=\lim_{x\rightarrow0} \frac{1+3x^2+o(x^2)}{1+o(x)}=\frac{1+0+0}{1+0}= 1$$

Answer 2

The left-hand side is $x^2+O(x^4)=x^2+o(x^3)$ because $x^4=o(x^3)$.

Answer 3

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