Let a function $f:X \times \mathbb{R} \rightarrow \mathbb{R}$ continuous, with $X \subset \mathbb{R}$ compact, and supose that $\partial\_2 f(x,t)$ exists for all $x \in X$ and is continuous. (here $\partial\_2$ is the derivative with respect to the second coordinate)


I would like to know if $$\displaystyle \lim\_{(x,t) \to (x\_0,0)} \frac{f(x,t)-f(x,0)}{t} = \partial\_2 f(x\_0,0)$$


I think it is not true, but i can't find a counterexample. Observe that if we take the limits separately (first $x\to x\_0$ and then $t \to 0$, or in the reverse order), these limits are both equal to $\partial\_2 f(x\_0,0)$. It seems to me like the classical examples of the functions whose limits in each coordinate exists separately but the total limit does not exist.


Can someone help me with this?


Thanks


EDITED (generalizing the question)


Supose the folowing situation


$f:X \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ continuous, with $X$ a compact metric space. Denote by $f\_x$ the function $f\_x(v) = f(x,v)$, and supose that for each $x \in X$, $f\_x: \mathbb{R}^m \rightarrow \mathbb{R}^m$ is differentiable and $$ (x,v) \mapsto (f\_x)'(v) \in Gl(\mathbb{R}^m)$$ is continuous.


Is it true that $$\lim\_{(x,v)\to (x\_0,0)} \frac{f(x,v)-f(x,0)}{\|v\|} = (f\_{x\_0})'(0)\cdot v ?$$



This limit converges to the partial derivative?
For $t\ne 0$ one has
$${f(x,t)-f(x,0) \over t}- \partial\_2 f(0,0)= \int\_0^1 (\partial\_2 f(x,\tau \thinspace \thinspace t) - \partial\_2 f(0,0))\thinspace d\tau ,$$
and here the right side is $<\epsilon$ when $(x,t)$ is in a suitable neighbourhood of $(0,0)$.


For an $f:X\times {\bf R}^n\to {\bf R}^m$ it is enough to consider the $i$-th coordinate function $f\_i:X\times {\bf R}^n\to {\bf R}$, again denoted by $f$, and for the latter consider the auxiliary function
$\phi(t):=f(x,t \thinspace v)$ on the interval $[0,1]$. One gets
$$f(x,v)-f(x,0) =\int\_0^1 \phi'(t)\thinspace dt = \int\_0^1 \nabla f\_2(x,t\thinspace v)\cdot v \thinspace dt,$$ whence 
$$f(x,v)-f(x,0) = \nabla f\_2(x\_0,0)\cdot v + o(\|v\|) \qquad ((x,v)\to(x\_0,0)).$$ 


