# A small part of a chain rule proof I don't get

• Jan 10th 2011, 10:22 AM
Stonehambey
A small part of a chain rule proof I don't get
Hello,

I've been trying to get my head round part of a proof and I don't see the following implication.

Suppose we have

$\lim_{x \to x_0} E(g(x)) = E(g(x_0)) = 0$

$\lim_{x \to x_0}[f'(g(x_0) + E(g(x)))]\frac{g(x) - g(x_0)}{x - x_0} = f'(g(x_0))g'(x_0)$?

Since doesn't the above require that $f'$ is continuous at $g(x_0)$, and if it does, why is this necessarily true?

Thanks :)

Stonehambey
• Jan 10th 2011, 12:33 PM
Drexel28
Quote:

Originally Posted by Stonehambey
Hello,

I've been trying to get my head round part of a proof and I don't see the following implication.

Suppose we have

$\lim_{x \to x_0} E(g(x)) = E(g(x_0)) = 0$

$\lim_{x \to x_0}[f'(g(x_0) + E(g(x)))]\frac{g(x) - g(x_0)}{x - x_0} = f'(g(x_0))g'(x_0)$?

Since doesn't the above require that $f'$ is continuous at $g(x_0)$, and if it does, why is this necessarily true?

Thanks :)

Stonehambey

There's too little context here, but from what I can tell you used the continuity of $f$ at $g(x_0)$ to conclude that $\displaystyle \lim_{x\to x_0}f(g(x))=f(g(x_0))$ it seems. Care to fill in the rest of the proof?
• Jan 11th 2011, 12:39 AM
Stonehambey
Hi,

The proof is from Trench's Introduction to Real Analysis, which can be freely downloaded here

William Trench - Trinity University Mathematics

I was wary of posting the entire proof verbatim in case of any infringement (not sure what the laws are concerning mathematical proofs but I thought it better to be safe).

The proof in question is on page 78 of the book.

Thanks,

Stonehambey

EDIT: Actually I believe there may be a bracket missing from equation (10) in the proof, immediately after $f'(g(x_0))$. Can anyone confirm this?
• Jan 11th 2011, 12:54 AM
Drexel28
Quote:

Originally Posted by Stonehambey
Hi,

The proof is from Trench's Introduction to Real Analysis, which can be freely downloaded here

William Trench - Trinity University Mathematics

I was wary of posting the entire proof verbatim in case of any infringement (not sure what the laws are concerning mathematical proofs but I thought it better to be safe).

The proof in question is on page 78 of the book.

Thanks,

Stonehambey

EDIT: Actually I believe there may be a bracket missing from equation (10) in the proof, immediately after $f'(g(x_0))$. Can anyone confirm this?

Ha! Luckily for you I own Trench, I don't think many people would download it :). Anyways the part you mention reads

$\displaystyle \frac{h(x)-h(x_0)}{x-x_0}=\left[f'\left(g(x_0)+E(g(x))\right]\frac{g(x)-g(x_0)}{x-x_0}\right]$

$\displaystyle \frac{h(x)-h(x_0)}{x-x_0}=\left[f'\overbrace{(g(x_0))}^{\text{here}}+E(g(x))\right]\frac{g(x)-g(x_0)}{x-x_0}\right]$