Is there a method such that one can determine the limit $\lim_{h \to 0}\frac{f(x)}{x^n}, n \in \mathbb{Z}^{+}$ for an implicitly defined function, defined near the origin - such as $2x^3-3x^2+2y^3+3y^2-y=0$ - given that the limit actually exists?
This post serves as a follow up to an answer given my previous question (see Limit of Implicitly Defined Function).
Richard P came with the astute observation that given the implicit function theorem tells us $y$ can be represented as a function of $f$, i.e. $y=f(x)$, one may use the definition of single variable differentiation and consider it at $x=0$, for:
\begin{align} f'(0)&=\lim_{h \to 0}\frac{f(0+h)-f(0)}{h} \\ &=\lim_{h \to 0}\frac{f(h)}{h} \end{align} as $f(0,0)=0$. Thus, all that one has to do is to use implicit differentiation, with respect to $x$, on $f(x,y)=0$, rearrange and then let $x \to 0$ to determine $\lim_{x \to 0}\frac{f(x)}{x}$.
If we know the given limit exists then, as $\;f(x)\;$ is assumed differentiable, it must be that $\;f(0)=0\;$ , so we can apply l'Hospitla's rule:
$$\lim_{x\to 0}\frac{f(x)}x=\lim_{x\to 0}f'(x)$$
But
$$2x^3-3x^2+2y^3+3y^2-y=0\implies6x^2-6x+(6y^2+6y-1)y'=0\implies$$
$$\implies f'(x)=y'=\frac{6x-6x^2}{6y^2+6y-1}$$
if the last expression's denominador isn't zero, and if $\;6y^2+6y\rlap{\;\;\;\;/}\xrightarrow[x\to 0]{}1\;$, then our limit is zero. Otherwise we may need more conditions to be sure of anything.