Let $f:\mathbb{R} \to \mathbb{R}$ be $C^3$. Show the equivalence: $f^{(k)}(0) = 0$ for $k=0,1,2 \iff \lim_{x\to 0} \frac{f(x)}{x^3}$ exists Let $f:\mathbb{R} \to \mathbb{R}$ be  $C^3$. Show the equivalence: 
$$f^{(k)}(0) = 0 \quad k=0,1,2 \iff \lim_{x\to 0} \frac{f(x)}{x^3} \text{, exists.}$$
Trying: 
Since $f \in C^3$, implies $f, f', f''$ are differentiable (therefore continuous) and $f'''$ is continuous in all domain.
Therefore we can approach $f$ with a order 3 Taylor polynomial, such as:
$$f(a+x) = f(a) + f'(a)x + \frac{f''(a)}{2}x^2 + r(x)$$
Where $\lim_{x \to 0} \frac{r(x)}{x^3} = 0$. (can I assume that? isn't what I'm trying to prove? This is the definition of Infinitesimal Taylor Polynomial. ).
$(\Rightarrow)$ Let's solve for $r(x)$
$$ r(x) = f(a+x) - f(a) - f'(a)x - \frac{f''(a)}{2}x^2 $$
Dividing by $x^3$
$$ \frac{r(x)}{x^3} = \frac{f(a+x)}{x^3} - \frac{f(a)}{x^3} - \frac{f'(a)}{x^3}x - \frac{f''(a)}{2x^3}x^2 $$
Lets take $a=0$
$$ \frac{r(x)}{x^3} = \frac{f(x)}{x^3} - \frac{f(0)}{x^3} - \frac{f'(0)}{x^3}x - \frac{f''(0)}{2x^3}x^2 $$
By hypothesis $f^{(k)}(0) = 0 \quad k=0,1,2$. Let's substitute:
$$ \frac{r(x)}{x^3} = \frac{f(x)}{x^3} - \frac{0}{x^3} - \frac{0}{x^3}x - \frac{0}{2x^3}x^2 $$
Taking the limit
$$ \lim_{x \to 0} \frac{r(x)}{x^3} = \lim_{x \to 0} \frac{f(x)}{x^3} - \lim_{x \to 0} \frac{0}{x^3} - \lim_{x \to 0} \frac{0}{x^3}x - \lim_{x \to 0} \frac{0}{2x^3}x^2 $$
Since $f$ is continuous
$$ \lim_{x \to 0} \frac{r(x)}{x^3} = \lim_{x \to 0} \frac{f(x)}{x^3} - 0 - 0 - 0 $$
$$ \lim_{x \to 0} \frac{r(x)}{x^3} = \lim_{x \to 0} \frac{f(x)}{x^3} $$
Therefore $ r(x) = f(x) $ since the limit of $ \lim_{x \to 0} \frac{r(x)}{x^3} $ exists (is a Taylor Poly) we conclude that lim of $ \lim_{x \to 0} \frac{f(x)}{x^3} $ also exists.
Is that correct? Also I don't have any idea in how to do the other direction.

Use the Taylor remainder theorem: for each $x > 0$ there exists $0 < c < x$ with the property that $$f(x) = f(0) + f'(0) x + \frac{f''(0)}{2}x^2 + \frac{f'''(c)}{6} x^3.$$ The stated hypotheses imply that $$\frac{f(x)}{x^3} = \frac{f'''(c)}{6}.$$ Write $c = c_x$ to emphasize the dependence on $x$. The definition of the limit and continuity of $f'''$ imply that $$\lim_{x \to 0+} \frac{f'''(c_x)}{6} = \frac{f'''(0)}{6}.$$ Thus $$ \lim_{x \to 0^+} \frac{f'''(x)}{x^3} = \frac{f'''(0)}{6}.$$ The proof for the left-sided limit is nearly identical.
