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"$\textit{$f$ is differentiable at $a$, then $f$ is continuous at $a$}$"

This is the statement proven in my lecture notes, however I think the proof is wrong.

Recall $$\lim_{x\to a} f(x) = f(a)$$ Then we rearrange and have $$\lim_{x\to a} (f(x)-f(a)) =\lim_{x\to a} \left(\frac{f(x)-f(a)}{x-a}\right)(x-a) = \lim_{x\to a} \left(\frac{f(x)-f(a)}{x-a}\right)\lim_{x\to a}(x-a) = f'(a)\cdot 0 = 0$$

Why is this a proof of the above statement? Okay I understand that at the end we obtain zero so we can rewrite the term on the LHS as the "definition" of continuity. However we started exactly from the "definition" of continuity. Indeed we started from $\lim_{x\to a} (f(x)-f(a))$ which by Arithmetic of Limits is equal to the above. Hence we wanted to prove $A \implies B$ but we started from $B$ and we showed it works for $A$. This is a circular reasoning!

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    okay my example below doesn't use exactly the same logic, but I hope what I mean is clear – 2017-01-23
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    Look at what it says backwards – 2017-01-23
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    It makes sense to write $\lim_{x\to a} (f(x)-f(a))$ even if $f$ is not continuous at $a$, in which cases you are writing a number that is not $0$ or something that is not a number. But solely writing the expression does not carry the assumption that $f$ is continuous at $a$. – 2017-01-23
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    @mathbeing oh right! So the whole thing then works because $f'(a)$ exists right? – 2017-01-23
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    Yes!! You start with that limit which you don't know even if it is a number. But you can express it in other way by performing some formal arithmetic and the result must be the same, even if you started with something that does not make sense. As you end up with $0$, the limit you didn't know anything about must be the $0$ number. – 2017-01-23
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    You are not multiplying both sides by 0 in the above proof. In the first equality you are multiplying by $\frac{x-a}{x-a}$ which goes to 1 as $x$ goes to $a$. – 2017-01-23

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better is we start from the Expression $$\frac{f(x)-f(a)}{x-a}\cdot (x-a)$$ and now you must ask under which conditions the Limit exists

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    This makes sense! Thank you! – 2017-01-23