Limit of $ \frac{f(h+k) - f(k) - f(h) + f(0)}{hk}$ Calculate the following limit when $(h,k)\to (0,0)$ when $f$ is twice differentiable in 0.
By removing $f(0) + xf'(0) + x^2f''(0)/2$ to $f$, we can suppose WLoG that $f(0)=f'(0) = f''(0)=0$. 
Let $g:h\mapsto f(h+k) -f(h) - f(k)$
Then $g$ is differentiable around 0 and $g'(h) = f'(h+k) - f'(h)$.
When $f$ is twice continuously differentiable around 0, I managed to conclude by writing the difference as an integral and using the fact that $f''(0) = 0$ to find an upper bound on $g'$ around 0.  So the limit here should be 0. So that in the general case we should  get $f''(0)$
But I am stuck in the general case ! When I try to find an upper bound on this derivative, I can only get something of the form $\varepsilon |h+k| + \varepsilon |h|$ and I cannot conclude with that...
Any tips ? 
 A: Your conjecture is correct:

Let $f:(-r, r) \to \Bbb R $ be differentiable, and assume that $f''(0)$ exists. Then
  $$ \tag{*}
 \lim_{(h, k) \to (0, 0)} \frac{f(h+k) - f(k) - f(h) + f(0)}{hk} = f''(0) \, .
$$

Remark: The continuity of $f''$ (or even the existence of $f''(x)$ for $x \ne 0$) is not needed for this conclusion.
The idea is to apply the mean value theorem to one of
$$
\bigl(f(h+k) - f(k) \bigr) - \bigl(f(h) - f(0) \bigr) \\
\bigl(f(h+k) - f(h) \bigr) - \bigl(f(k) - f(0) \bigr) \\
$$
depending on whether $|h| \le |k|$ or $|k| \le |h|$.
Proof: As you already said we can assume that $f'(0) = f''(0) = 0$, so that
$$
 \lim_{x \to 0} \frac{f'(x)}{x} = 0 \, .
$$
Let $\epsilon > 0$. There is a $\delta > 0$ such that $|f'(x) | \le \epsilon |x|$ if $|x| < \delta$. We will now show that
$$ \tag{**}
|f(h+k) - f(k) - f(h) + f(0)| \le 3 \epsilon |hk|
$$
if $|h| + |k| < \delta$, and that implies $(*)$.
Since $(**)$ is symmetric in $(h,k)$ we can assume that $|h| \le |k|$. Then the mean value theorem gives
$$
f(h+k) - f(k) - f(h) + f(0) = hf'(k+ \theta_1 h) -hf'(\theta_2h)
$$
with $\theta_1, \theta_2 \in (0,1)$. It follows that
$$
|f(h+k) - f(k) - f(h) + f(0)| \le |hf'(k+ \theta_1 h)| +|hf'(\theta_2h)| \\
\le \epsilon |h|(|k+ \theta_1 h | + |\theta_2 h|) \\
\le \epsilon |h| (|k| + 2 |h|) \le  \epsilon |h| \cdot 3|k|
$$
and that completes the proof.
