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This Proof of Chain Rule comes from the textbook of Stewart

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I'm stuck here: " If we define $\epsilon$ to be $0$ when $\Delta x=0$, then $\epsilon$ becomes a continuous function of $\Delta x$".

How come $\epsilon$ becomes a continuous function of $\Delta x$? $\quad$ Isn't $\epsilon$ just continuous at $\Delta x=0$?

Thank you so much!

Andrew Li
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    You are absolutely right, except if somewhere earlier in the text it was specified that $f$ is continuous. Then $\varepsilon$ would also be continuous. It makes no difference for the chain rule, though. – Vercassivelaunos May 26 '23 at 10:33
  • I prefer to write $f(a+h)=f(a)+f'(a)\cdot h+o(h)$, or $f(a+h)=f(a)+f'(a)\cdot h+\psi(h)$ where $\lim_{h\to0}\psi(h)/h=0$ – FShrike May 26 '23 at 10:48

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if $\epsilon$ is continuous in $\Delta x$ then $\epsilon$ is bounded in any closed interval around $0$:

$I=[-\Delta x, \Delta x]$ this prevents $\epsilon$ from becoming infinite when $\Delta x=0$.

and the limit exists.

ryaron
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