Prove that if $f$ is differentiable in $\mathbb{R}^n$ then it is continuous. Prove that if $f$ is differentiable in $\mathbb{R}^n$ then it is continuous.
I have found a proof online but am having a hard time following some of the steps. 
$||f(x+h) - f(x)|| -||L(h)|| < \epsilon ||h||$
Okay, this makes sense still, then
$||f(x+h) - f(x)|| < \epsilon||h||+||L(h)||< \epsilon||h||+||h|| ||Df(x)||$
I am not exactly sure what $Df(x)$ represents but I am assuming it is the directional derivative. If so, why is this inequality true?? 
 A: $Df(x)$ is a linear function, it takes each vector $v\in\mathbb{R}^n$ in it's directional derivative $Df(x)\cdot v = \frac{\partial f}{\partial v}(x)$. The norm $\| Df(x)\|$ usually is defined as $\sup\{\|Df(x)\cdot v\|_2; \ v\in\mathbb{R}^n,\|v\|_2=1\}$.
A: $Df(x)$ is probably the differential of $f$ at $x$, the confusing thing is that the notation is the same for the norm of vectors and the norm of applications, $\lVert Df(x)\rVert$ is often noted $\lvert\lVert Df(x)\rvert\lVert$ to make it easier to read. It is the operator norm for $\mathcal{L}(E)$ where $E$ is a normed vector space and defined by $$\vert\lVert φ\rvert\rVert=\sup_{\lVert a\rVert⩽1}\lVert φ(a)\rVert$$
For reference, here is a proof that doesn't use the operator norm.
Using Landau notation, if $f$ is differentiable in $a$, we have by definition for all $h∈ℝ^n$,
$$f(a+h)=f(a) + df(a)(h) + ο_{h→0}(h)$$
where $df(a)$ is a countinuous linear application.
So we have
$$\lVert f(a+h)-f(a)\rVert=\lVert df(a)(h) + ο_{h→0}(h)\rVert$$
but $df(a)$ is linear, so $df(a)(0)=0$; and continuous, so $df(a)(h)\longrightarrow 0$ when $h→0$. Thus
$$df(a)(h) + ο_{h→0}(h) \longrightarrow 0$$ when $h→0$.
Therefore $$\lVert f(a+h)-f(a)\rVert \longrightarrow 0$$ when $h→0$, i.e. $f$ is continuous at $a$.
So if $f$ is differentiable over any open subset $O$ of $ℝ^n$, it is continuous over $O$.
Note that this proof doesn't even need finite dimension, it is just a direct consequence of the definition of differentiability.
