Derivatives Theorems for functions from $\mathbb{R}$ to a Banach Space The following are true for functions $f$ from $\mathbb{R}$ to $\mathbb{R^n}$:
Suppose f is continuous, $f'$ exists in $(x-\delta,x) \cup (x,x+\delta)$ and 
$\lim\limits_{u \to x}{f'(u)} = l$ then $f'(x)=l$
Suppose $f'$ is continuous on $[a,b]$ then 
$\forall{\epsilon >0}$ $\exists \delta>0$ s.t.
$\forall x,t \in [a,b] $, $|x-t|<\delta \implies 
|{\frac{f(t)-f(x)}{t-x} - f'(x)}| < \epsilon$
Are they true for functions from $\mathbb{R}$ to a Banach space as well?
These are from Baby Rudin Ch 5 Q 8,9, but he doesn't dicuss any generalizations. I'm not very familiar with theorems on Banach spaces, so I apologize if this is standard.
 A: The MVT also applies with values in a Banach space $E$ (even with the argument in a Banach space, but we leave this aside). If $g:[a,b]\rightarrow E$ is continuous and if it is differentiable at every point in $(a,b)$ then:
  $$ |g(b)-g(a)|_E \leq \left( \sup_{a<t<b} |g'(t)|_E \right) |b-a|$$
If wanted I can provide a short proof.
For your first claim: We assume that given $\epsilon>0$ there is $0<r<\delta$ so that 
$$0<|u-x|<r \Rightarrow |f'(u)-\ell|_E <\epsilon.$$
For such $u$ we may apply the MVT on $g(t)=f(t)-f(x)-\ell(t-x)$, $g'(t)=f'(t)-\ell$  to conclude that
    $$ |f(u)-f(x)-\ell(u-x)|_E = |g(u)|_E = |g(u)-g(x)|_E \leq \epsilon |u-x|$$
The second claim is a consequence of $[a,b]$ being compact, so $f'$ continuous on $[a,b]$ implies that it is uniformly continuous (true also for values in a Banach space $E$). So given $\epsilon>0$ there is $\delta>0$ so that
$$ x,t\in [a,b] : |x-t|<\delta \Rightarrow |f'(t)-f'(x)|_E < \epsilon$$
By the mean value theorem we get for such $x,t$ (below $(x;t)$ denotes the segment between $x$ and $t$):
$$ |f(t)-f(x) -f'(x)(t-x)| \leq \left(\sup_{y\in (x;t)} |f'(y)-f'(x)|\right) |t-x| $$
and the claim follows since the value in the paranthesis is $<\epsilon$.
A: Here is a proof for part 1, actually quite straightforward.
First assume $l=0$. Given $\epsilon>0$ let $\delta'$ be s.t. for each $u \in (x - \delta',x+\delta')$ , $u \neq 0$ we have $|f'(u)|<\epsilon$
Then ${{|f(u)-f(x)|} \over {|u-x|}} \leq \sup_{t \in (u,x)} {|f'(t)|} \leq \epsilon$ for each $u \in (x - \delta',x+\delta')$ , $u \neq 0$
Hence $f'(x)=0$
Now if $l \neq 0$ we let $g(x)=f(x)-xl$ apply the result and the general case follows.
