I want to prove the following statement:

Let $I$ be some interval in $\mathbb{R}$, that is $I \subset \mathbb{R}$

Let $f:I \rightarrow \mathbb{R}^n$ be differentiable in $a$. Then $f$ is continuous in $a$. 

Proof:

Notice that $f(x)-f(a)=(x-a) \frac{f(x)-f(a)}{x-a}$. 

We can restrict the function to $I \setminus \{a\}$ and call it $f\_0$. 

Then notice that:
$\lim\_{x\to a}f\_0(x) -f(a)=\lim\_{x\to a}(x-a)· \lim\_{x\to a}\frac{f\_0(x)-f(a)}{x-a}= 0 · f'(a)=0$, by using the above. 

So we conclude that $\lim\_{x\to a}f\_0(x) =f(a)$. 

Now how do I conclude from this that 

$\lim\_{x\to a}f(x) =f(a)$?