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.
 A: METHOD 1: Use L'Hospital's Rule
L'Hospital's Rule states that if the limit 
$$\lim_{x\to0}\frac{f'''(x)}{1}$$
exists and $\lim_{x\to 0}f''(x)=0$ and $\lim_{x\to 0}x=0$, then 
$$\lim_{x\to 0}\frac{f''(x)}{x}= \lim_{x\to0}\frac{f'''(x)}{1}$$
Since we are given that $\lim_{x\to0}\frac{f'''(x)}{1}$ exists, along with $\lim_{x\to 0}f''(x)=0$, and it is trivially true that $\lim_{x\to 0}x=0$, then we have recursively 
$$\lim_{x\to0}\frac{f(x)}{x^3}=\lim_{x\to0}\frac{f'(x)}{3x^2}=\lim_{x\to0}\frac{f''(x)}{6x}=\frac16\lim_{x\to0}f'''(x)$$
and thus, $\lim_{x\to0}\frac{f(x)}{x^3}$ exists.

METHOD 2: Use the extension of the mean value theorem
If you would rather use the Extended Mean Value Theorem, or Taylor's Formula with a Remainder, then we know that there exists a $\xi$, $0<\xi<x$, such that 
$$f(x)=f(0)+f'(0)x+\frac12 f''(0)x^2+\frac16 f'''(\xi)x^3$$
Thus,
$$\lim_{x\to 0}\frac{f(x)}{x^3}=\lim_{x\to 0}\frac16 f'''(\xi)=\frac16 f'''(0)$$
which exists by hypothesis. Note that the last equality is obvious from the continuity of $f'''$.  
However, as pointed out by @ErickWong, there is a stronger form of Taylor's Theorem with a Remainder attributed to Peano, that states if $f, f', cdots ,f^{(k)}$ exist at a point $x_0$, then there is a function $R_k$ such that 
$$f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac12 f''(x_0)(x-x_0)^2+\cdots + f^{(k)}(x_0)(x-x_0)^k+R_k(x)(x-x_0)^k$$
and thus $f/x^3\to \frac16f'''(0)$ even if $f'''$ is not continuous at $0$
with $\lim_{x\to x_0}R_k(x)=0$.
A: 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.
A: Lemma: Suppose $g\in C^2(\mathbb {R})$ and $g'''(0)$ exists. If $0=g(0)=g'(0)=g''(0)=g'''(0),$ then $g(x) = o(x^3)$ as $x\to 0.$ Proof:
$$g(x) = g(x)-g(0) = g'(c_x)x = (g'(c_x)-g'(0))x= g''(d_x)c_xx= (g''(d_x)-g''(0))c_xx = [(g''(d_x)-g''(0))/d_x]d_xc_xx$$
where we've used the MVT twice. Since $c_x,d_x$ are in between $0$ and $x,$ and the term in brackets $\to g'''(0)=0,$ we see the above is $o(x^3)$ as desired.
Now consider an arbirary $f\in C^2$ such that $f'''(0)$ exists. Let $P(x) = f(0)+f'(0)x + f''(0)x^2/2 + f'''(0)x^3/3!.$ Then by the lemma,
$$f(x) = P(x) + o(x^3).$$
The conclusion in this problem for $f$ follows easily.
