Missing step in differentiability implies continuity.

Question

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

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

Answer 1

1) Whenever $x=a$, $|f(x)-f(a)|=|f(a)-f(a)|<0<\epsilon$.

2)For all other values of $x$ in the domain, we notice that $f_0(x)$ converges to $f(a)$ as $ x \rightarrow a $.

With these two facts combined we can say that whenever $x \rightarrow a$, we notice that $f(x) \rightarrow f(a)$. Consequently, $f(x)$ is continuous.

Answer 2

By definition of differentiability for each component we have that

$$f_i(a+h)=f_i(a)+f'_i(a)\cdot h + o(h)$$

then taking the limit both sides

$$\lim_{h\to 0} f_i(a+h)=\lim_{h\to 0}f_i(a)+f_i(a)\cdot h + o(h)=f_i(a)$$

that is

$$\lim_{x\to a} f_i(x)=f_i(a)$$

then

$$\lim_{x\to a} f(x)=f(a)$$