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 $$\lim_{(x,v)\to (x_0,0)} \frac{f(x,v)-f(x,0)}{\|v\|} = (f_{x_0})'(0)\cdot v ?$$

`$\partial_2f(x,u)$`

for some $u$ between 0 and $t$. Since you're assuming continuity of this partial derivative, it will approach`$\partial_2f(x_0,0)$`

. (I'm assuming that your continuity assumption for the partial derivative means continuity with respect to both variables together, not some sort of separate continuity.) – Andreas Blass Feb 11 '11 at 19:26