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Let $f:(a\,..b)\rightarrow \mathbb R$ or $f:(a\,..b) \rightarrow \mathbb R$ be a function differentiable on $x_0 \in (a\,..b)$ or $x_0 \in (a\,..b)$.

I'm now reading a proof that all differentiable functions are continuous.

This proof assumes that $\displaystyle \lim_{x\to x_0} \big(f(x)-f(x_0)\big)$ exists.

How to prove this?

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    Diifferentiable implies continuous. – 2017-02-26
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    @ZacharySelk, that's certainly true but that's what OP is trying to prove. – 2017-02-26
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    @MeesdeVries I know, that's why it's a comment. :) Just something for them to google – 2017-02-26

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The proof in your link is better if you read it bottom to top; that is, $$ 0 = f'(x_0) \times 0 = \lim_{x \to x_0} \frac{f(x_0) - f_(x)}{x_0 - x} \times \lim_{x \to x_0} (x_0 - x) = \lim_{x \to x_0} \left(f(x_0) - f(x)\right). $$ Then, the step that happens in the final equality is "if two limits exist, then so does the limit of their product and it is equal to the product of the limits". The "converse" of that doesn't hold.