Consider the equation $2x^3-3x^2+2y^3+3y^2-y=0$. It is possible to show, using the implicit function theorem, that this defines a function $y=f(x)$ in a neighborhood of $(0,0)$ [see my reasoning below]. Given this, determine the limit of $\frac{f(x)}{x}$ as $x \to 0$.
I must admit I cannot think of any useful thing (for the moment) to do in order to determine the limit.
$y=f(x)$ is valid in the neighborhood of $(0,0)$
The interpretation of the implicit function theorem (Rudin, pp. 224-225) when $n=m=1$ is as follows (my response to problem 20, p. 241):
If $f(x,y)$ is continuously differentiable in a neighborhood of $(x_0,y_0)$, $f(x_0,y_0)=0$, and $D_2(x_0,y_0) \not =0$, then there exist an interval $I=(x_0- \delta,x_0 +\delta),$ an interval $J=(y_0-\eta, y_0 +\eta) $, and a continuously differentiable function $\phi:I \to J$ such that for all $(x,y) \in I \times J$ the equation $f(x,y)=0$ holds if and only if $y = \phi (x)$.
We note $f(x,y)$ is a continuously differentiable real function in the plane, $f(0,0)=0$ and $\frac{\partial f}{\partial y}(0,0) =-1 \not =0$. As such, it satisfy the assumptions of the implicit function theorem, and so it is possible to write $y=f(x)$ in the neighborhood of $(0,0)$.
Best Answer
Through the implicit function theorem, we are guaranteed that y can be represented as a single variable function of x, this helps a lot with finding that limit.
Recall that $f(0,0)=0$ and think back to the one of definitions of a single variable derivative:
$$\frac{df}{dx}=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$
Now, if we consider the derivative at x=0 we get the following:
$$f'(0)=\lim_{h \to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h \to 0}\frac{f(h)}{h}$$
So, all we have to do is find the derivative of $f(x)=y$ with respect to $x$ which is simply done by implicitly differentiations the original function in x.