I am trying to prove or disprove the next statement that seems necessary for the proof of Proposition 2.9 of this [book](https://books.google.de/books?id=e8tXjbBGEroC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false).



> 
> Let $U\subset R^k$ be compact and $f:R^n\times U \to R^m$ be twice differentiable with respect to $x\in R^n$. Assume that $f$ and the derivatives are continuous with respect to $(x,u) \in R^n\times U$. Let $\lambda\_i > 0$ and $u\_i \in U$, $i=1,2,3,\dots$, such that $\lambda\_i \to 0$ and $u\_i \to u\_0\in U$ as $i\to\infty$. Then,
>  \begin{equation}
> \tag{1}\label{1}
> \lim\_{i\to\infty}\lambda\_i^{-1}\left|f(\hat{x}+\lambda\_ix,u\_i) - f(\hat{x},u\_i) - D\_xf(\hat{x},u\_i)(\lambda\_ix)\right| = 0
> \end{equation}
>  where $D\_xf(\hat{x},u\_i)\in L(R^n,R^m)$ is the derivative of $f$ with respect to $x$ at $(\hat{x},u\_i)$.
> 
> 
> 


\eqref{1} is obvious if $u\_i = u\_0$, $i = 1,2,\dots$, by the definition of the derivative but seems not trivial for me in other case. Would you give me any hint or a reference?



Proof of $\lim_{i\to\infty}\lambda_i^{-1}\left|f(\hat{x}+\lambda_ix,u_i) - f(\hat{x},u_i) - D_xf(\hat{x},u_i)(\lambda_ix)\right| = 0$
By Taylor's theorem with remainder, 


$$ | f(\hat{x} + \lambda\_i x, u\_i - f(\hat{x},u\_i) - D\_x f(\hat{x},u\_i)(\lambda\_i x) | \leq C \lambda\_i^2 |x|^2 \sup\_{y\in B\_{\lambda\_i |x|} (\hat{x})} |D\_x^2 f(y,u\_i)| $$


for some universal constant $C$. For sufficiently large $i$, the right hand side is bounded by 


$$ \sup\_{y\in B\_1(x), u\in B\_1(u\_0)} |D^2\_xf(y,u)| $$


the claim then follows as $\lambda\_i \searrow 0$. 


