0
$\begingroup$

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

  • 0
    Sorry, I just added the link.2012-05-25

2 Answers 2

1

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

0

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.

  • 0
    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.2012-05-25