Differential Calculus - Swapping a sup and a limit. Well, I'm doing some exercises on differential calculus and I'm stuck.
(a) Let $U \subset \mathbb R^m$ and $f: U \rightarrow \mathbb R^n$ be a continuous function on a line segment $[x, x+h]\subset U$ and differentiable on $]x, x+h[$. Show that if $T:R^m\rightarrow R^n$ is a linear map, then:
$$ ||f(x+h)-f(x)-T(h)\leq \sup_{t \in ]0, 1[}||df(x+th)-T||\,||h|| $$
(b) Let $U \subset R^m$ and $f:U\rightarrow \mathbb R^n$ a continuous function differentiable on $U- \{x\}$, where $x$ is a interior point of $U$. Suppose $\lim_{y \rightarrow x}df(y)=T$ for some linear map $T: \mathbb R^m \rightarrow \mathbb R^n$. Show that $f$ is differentiable on $x$ and $df(x)=T$. Suggestion: Use item (a).
Progress:
I had no problems with (a) but I'm stuck at be. I must show that:
$$\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)-T(h)}{||h||}=0$$
We know that:
$$\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)-T(h)}{||h||}\leq \lim_{h \rightarrow 0}\sup||df(x+h)-T||$$
If I can swap the sup and the limit then I'm done, but I don't know if I'm allowed to.
Ps: I don't know why, but I cant tag this question properly.
 A: Hint: Let $\epsilon > 0$ be arbitrary and $\{h_n\}$ be any sequence $0 \neq h_n \to 0$. Then you can find $t_n \in \,]0,1[$ such that $\|f(x+h_n)-f(x)-Th_n\| \leq \big(\|df(x+t_nh_n) - T\|+\epsilon\big)\|h_n\|$ (why can we find such a $t_n$?). Deduce that $t_nh_n$ also $\to 0$ and then use this to conclude that 
$$
\overline{\lim_{n \to \infty}} \frac{\|f(x+h_n) - f(x) -Th_n\|}{\|h_n\|} \leq \epsilon.
$$ 
Since $\epsilon$ and $\{h_n\}$ were chosen arbitrarily, you'll be done after this!
A: Let $\epsilon>0$ be given and let $h_n$ be an arbitrary sequence such that $h_n\neq 0$ and $h_n \rightarrow 0$.
For each $n$, there exists a $t_n \in ]0, 1[$ such that $$\|f(x+h_n)-f(x)-Th_n\|\leq(\|df(x+t_n h_n)-T\|+\epsilon)\|h_n\|$$
(Otherwise, we would have that $\forall t\in (0, 1), \, \|f(x+h_n)-f(x)-Th_n\|>(\|df(x+t h_n)-T\|+\epsilon)\|h_n\|$, therefore  $ \|f(x+h_n)-f(x)-Th_n\|\geq(\sup_{t \in (0, 1)}\|df(x+t h_n)-T\|+\epsilon)\|h_n\|$ $>\sup_{t \in (0, 1)}\|df(x+t h_n)-T\|\|h_n\|$, which is absurd).
Since $t_n$ is bounded, $x+t_n h_n\rightarrow x$ as $n\rightarrow \infty$. Therefore, by our hypothesis, $\limsup\frac{\|f(x+h_n)-f(x)-Th_n\|}{\|h\|}\leq \epsilon $. Since $\epsilon$ is arbitrary, $\limsup\frac{\|f(x+h_n)-f(x)-Th_n\|}{\|h_n\|}\leq 0 $. But $\liminf\frac{\|f(x+h_n)-f(x)-Th_n\|}{\|h_n\|}\geq 0 $, therefore $\lim_{n\rightarrow \infty}\frac{\|f(x+h_n)-f(x)-Th_n\|}{\|h_n\|}= 0 $. We conclude that $\lim_{n\rightarrow \infty}\frac{f(x+h_n)-f(x)-Th_n}{\|h_n\|}= 0 $, and since $h_n$ was arbitrary, we have shown that $\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)-Th}{\|h\|}= 0$. $\square$
