# Squeeze theorem with differentiation

• February 23rd 2011, 07:13 AM
Dagger2006
Squeeze theorem with differentiation
I've attached the problem typed up to be more clear. I'm basically having trouble on how to prove that g'(x) exists using only that I know that f(x)<=g(x)<=h(x) for some neighborhood around x_0. I'm pretty sure I have to use some definition of the differentiation, but can't quite prove to myself that it exists without having more information.

Any help would be appreciated.

Attachment 20905
• February 23rd 2011, 07:30 AM
TheEmptySet
Quote:

Originally Posted by Dagger2006
I've attached the problem typed up to be more clear. I'm basically having trouble on how to prove that g'(x) exists using only that I know that f(x)<=g(x)<=h(x) for some neighborhood around x_0. I'm pretty sure I have to use some definition of the differentiation, but can't quite prove to myself that it exists without having more information.

Any help would be appreciated.

Attachment 20905

First note that by the squeeze theorem that

$\displaystyle \lim_{x \to x_0} f(x)\le \lim_{x \to x_0} g(x) \le \lim_{x \to x_0} h(x)$ but since $f(x_0)=h(x_0)$ they must also equal $g(x_0)$

Now in any neighborhood of $x_0$ you have

$\displaystyle f(x_0+k) \le g(x_0+k) \le h(x_0+k)$

Now use the fact above that $f(x_0)=h(x_0)=g(x_0)$ are equal to get

$\displaystyle f(x_0+k)-f(x_0) \le g(x_0+k)-g(x_0) \le h(x_0+k)-h(x_0)$

Can you finish from here?
• February 23rd 2011, 07:36 AM
Prove It
If $\displaystyle f(x) \leq g(x) \leq h(x)$ in some neighbourhood of $\displaystyle x_0$, then

$\displaystyle f(x_0 + \Delta x) \leq g(x_0 + \Delta x) \leq h(x_0 + \Delta x)$, where $\displaystyle \Delta x$ is some small change in $\displaystyle x$ in the neighbourhood of $\displaystyle x_0$...

$\displaystyle \frac{f(x_0 + \Delta x)}{\Delta x} \leq \frac{g(x_0 + \Delta x)}{\Delta x} \leq \frac{h(x_0 + \Delta x)}{\Delta x}$

$\displaystyle \lim_{\Delta x \to 0}\frac{f(x_0 + \Delta x)}{\Delta x}\leq \lim_{\Delta x \to 0}\frac{g(x_0 + \Delta x)}{\Delta x} \leq \lim_{\Delta x \to 0}\frac{h(x_0 + \Delta x)}{\Delta x}$

$\displaystyle f'(x_0) \leq g'(x_0) \leq h'(x_0)$.

I believe this proves $\displaystyle g'(x_0)$ exists...
• February 23rd 2011, 07:37 AM
Prove It
Quote:

Originally Posted by TheEmptySet
First note that by the squeeze theorem that

$\displaystyle \lim_{x \to x_0} f(x)\le \lim_{x \to x_0} g(x) \le \lim_{x \to x_0} h(x)$ but since $f(x_0)=h(x_0)$ they must also equal $g(x_0)$

Now in any neighborhood of $x_0$ you have

$\displaystyle f(x_0+h) \le g(x_0+h) \le h(x_0+h)$

Now use the fact above that $f(x_0)=h(x_0)=g(x_0)$ are equal to get

$\displaystyle f(x_0+h)-f(x_0) \le g(x_0+h)-g(x_0) \le h(x_0+h)-h(x_0)$

Can you finish from here?

Maybe you should use a different symbol to represent the change in $\displaystyle x$, since $\displaystyle h$ is the name of a function...
• February 23rd 2011, 07:39 AM
TheEmptySet
You are absolutely correct. Thanks :)
• February 23rd 2011, 07:59 AM
Dagger2006
Thanks for all the help. You guys are stating pretty much as far as I got. My worry is that in the final line where you take the limit of everything. Does the limit of the g function being between the f and h functions enough to prove that it exists? I might be over thinking this, but just proving that the limit, if it exists, would be between the f'(x) and h'(x) might not be enough to actually prove that it exists? Or would it?

More specifically in terms of slope. You can have a undefined slope between a very steep positive slope and a very steep negative slope?

Thanks again for all the help and fast feedback!