# Spivak's proof of the chain rule in $\mathbb{R^n}$

Spivak's proof of the chain rule in $\mathbb{R^n}$. The proof can be found on page 19.

I'm confused by the last step. "Equation 6 now follows easily."

Page 19

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Could you kindly add the proof or give a link where the proof can be found? – user17762 May 25 '12 at 5:39
Spivak has more than one book... – copper.hat May 25 '12 at 5:40
Sorry, I just added the link. – Fred May 25 '12 at 5:42

I thought that I took this proof from Spivak. You might find it more clear.

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+1 for the great teacher and the awesome rap musician! – Georges Elencwajg May 25 '12 at 7:11

Divide across by $|x-a|$ giving $\frac{|\psi(f(x))|}{|x-a|} \leq \epsilon \frac{|\phi(x)|}{|x-a|}+\epsilon M$. Line (4) shows that the $\frac{|\phi(x)|}{|x-a|} \to 0$, hence bounded, so this shows that $\frac{|\psi(f(x))|}{|x-a|} \to 0$, since $\epsilon >0$ was arbitrary.

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Is it certain that $\lim_{x\ \to a}\frac{|\psi(f(x))|}{|x-a|}$ exists? – Fred May 25 '12 at 5:51
Yes, the above shows that the limit is $0$. That is, $\forall \epsilon>0$, $\exists \delta > 0$, such that when $|x-a| < \delta$, then $\frac{|\psi(f(x))|}{|x-a|} < \epsilon$. – copper.hat May 25 '12 at 6:02