I have a question about whether I am even close to correct.
Let $\mathbb{I}$ and $\mathbb{J}$ be open intervals, and the functions $f:\mathbb{I} \to R$ and $h:\mathbb{J}\to R$ have the property that $h(\mathbb{J}) \subset \mathbb{I}$, so the composition $f\circ h :\mathbb{J} \to R$ is defined. Show that if the following conditions are true
- $x_o$ is in $\mathbb{J}$
- $h$ is continuous at $x_o$
- $h(x)\neq h(x_o)$ if $x \neq x_o$
- $f$ is differentiable at $h(x_o)$
then
$\lim_{x\to x_0} \frac{f(h(x)) - f(h(x_o))}{h(x) - h(x_o)} = f'(h(x_o)).$
Proof I found:
Let $y = h(x)$ and $y_o = h(x_0)$. Then $\lim_{x\ \rightarrow x_o} \frac{f(h(x)) - f(h(x_o))}{h(x) - h(x_o)} = \lim_{y\ \rightarrow y_o} \frac{f(y) - f(y_o)}{y - y_o}.$
But by continuity of $h$, as $x$ $\rightarrow$ $x_o$, $h(x)$ $\rightarrow$ $h(x_o)$ or $y$ $\rightarrow$ $y_o$. Using this fact and the differentiability of $f$, we have
$\lim_{y\ \rightarrow y_o} \frac{f(y) - f(y_o)}{y - y_o} = f’(y_o) = f’(h(x_o)).$
$Q.E.D.$
This seems vastly too simple compared to the torrent of information given in the question, but if this proof works, that brings me to my second question related to the question immediately following this one. It says “Use exercise 6 to show that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is differentiable at $x_o$ = 1, then...” and I will post a picture because I do not wish to retype it all with LaTeX. http://oi47.tinypic.com/2nb93zb.jpg
Basically, I struggled to understand the first question but when I finally did, the solution seemed trivial and I suspect this second question will be the same way, but I have no idea what they are asking for, nor do I see how I can use this first proof. I would greatly appreciate any insight.