Definition: A function $f:A\subset R^n\to R^m$ is said to be differentiable at a point $a\in A$ if there exists a real $m\times n$ matrix $B$ such that the following equality holds: $$ \lim_{h\to 0}\frac{f(a+h)-f(a)-Bh}{|h|} = 0 $$
I'm using the usual definitions of limit and metric on $R^n$. I have the following theorem with it's demonstration:
Theorem: If $f:A\subset R^n\to R^m$ is differentiable at $a\in A$, then $f$ is continuous at $a$.
Proof: We can write $$f(a+h)-f(a) = |h|\left[ \frac{f(a+h)-f(a)-Bh}{|h|}\right]+Bh$$ so $f(a+h)-f(a)\to 0$ as $h\to 0$ therefore $f$ is continuous at $a$.
My aim is to rewrite the demonstration using epsilons and deltas, especially the implication denoted by the bolded word in the box above. So I started this way (I'll use $e$ and $d$ instead of greek letters for short):
Given $e>0$, we have that there exists $d>0$ such that
$$
0<|h|<d \implies \left|\frac{f(a+h)-f(a)-Bh}{|h|}\right|<e
$$
then we can write
$$
|f(a+h) - f(a)|= \left||h|\frac{f(a+h)-f(a)-Bh}{|h|}+Bh\right|
$$
$$
\le |h|\left|\frac{f(a+h)-f(a)-Bh}{|h|}\right| + |Bh|
$$
But I don't know what to do next to bound this last expression, because of the terms $|h|$ and $|Bh|$. Any help will be appreciated.
For small $d>0$, one has $0<|h|<d$ implies $\left|f(a+h)-f(a)-Bh\right|/|h|<1$, and $|h|<(1+\|B\|)^{-1}\epsilon$, so $|f(a+h)-f(a)|<|h|+|Bh|\leq|h|+\|B\||h|=(1+|B|)|h|<\epsilon$.