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I am learning Calculus. I have read a proof from MIT OpenCourseware on how differentiability implies continuity. Here's the link (https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-5-discontinuity/MIT18_01SCF10_Ses5e.pdf).

I get the proof. Or rather, I understand why all the steps of the proof are true.

But I don't understand why all these steps prove that a function that is differentiable is continuous. It seems to me the author just took the derivative equation/limit, then rearranged it into the continuity definition.

Can someone explain to me, how being able to rearrange the derivative to form the continuity definition proves that differentiability = continuity? Can this be done without epsilon delta proofs and any knowledge of proof theory? Thank you!

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  • $\begingroup$ The following is the proof in your link: Let $a_n, b_n$ be convergent sequences in $\Bbb R$. A usual result of analysis courses is that $a_n\cdot b_n$ also converges, with $\lim_n(a_nb_n)=\lim_n a_n\lim_n b_n$. Now let $x_n$ be an arbitrary sequence converging to $x$, it follows with $a_n=\frac{f(x_n)-f(x)}{x-x_n}$, $b_n=x-x_n$ that $a_nb_n$ converges to $f'(x)\cdot0=0$. But $a_nb_n=f(x_n)-f(x)$ and thus $f(x_n)-f(x)$ must converge to zero. Since this holds for an arbitrary sequence $x_n\to x$ you recover continuity of $f$ at $x$. $\endgroup$ – s.harp Sep 2 '18 at 14:39
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    $\begingroup$ "proves that differentiability = continuity" -- it doesn't; differentiability implies continuity, but not vice versa. $\endgroup$ – joriki Sep 2 '18 at 15:34

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