# Need some constructive criticism...(advanced calc proof)

• May 13th 2008, 05:06 PM
xifentoozlerix
Need some constructive criticism...(advanced calc proof)
I am unsure of this proof, and I would appreciate any input about whether my reasoning is on the right track. Due tomorrow btw.

The question is:

Suppose that $f$ is continuous everywhere, $f'$ exists for $x\neq0$, and $\mathop {\lim }\limits_{x \to 0} f'\left( x \right)$ exists. Prove that $f'\left( 0 \right)$ exists.
Professor said to use L'Hôpital's rule if we get stuck.

I started by trying to let $g\left( x \right) = f'\left( x \right)$ and working through an argument using L'Hôpital's rule, but I made some extra assumptions and such which dont seem to be reasonable. Any nudges in the right direction would be great. My work is pretty useless...and ultimately it seems like I wouldnt be proving the right thing anyway. I feel like I am overthinking (as usual) what would be an otherwise simple and elegant proof. If anyone really wants to see my attempt, I will post it, but it is almost certainly incorrect. Thanks in advance.

edit: I know intuitively that it must exist, since $f$ is continuous and $f'$ exists everywhere but $0$ (for sure). And since the two-sided limit exists at $0$, either $f'$ is continuous at $0$ or has a removable discontinuity there, but it must be defined there because $f$ is continuous at $0$ and does not have a corner there. There cannot be a corner at $f \left( 0 \right)$ because the two-sided limit exists, so we know that the derivative must exist there (as $f$ is continuous without a corner)... but I do not see how L'Hôpital would come into play in a formal proof?
• May 13th 2008, 07:48 PM
ThePerfectHacker
Quote:

Originally Posted by xifentoozlerix
Suppose that $f$ is continuous everywhere, $f'$ exists for $x\neq0$, and $\mathop {\lim }\limits_{x \to 0} f'\left( x \right)$ exists. Prove that $f'\left( 0 \right)$ exists.

The first step that I like to do a lot of times is to formally state the problem. I find that it helps:
"Let $f:\mathbb{R}\mapsto \mathbb{R}$ be continous and let $f$ be differenciable on $(-\infty, 0)\cup (0,\infty)$ (possibly at $x=0$ also, but we do not know this). Suppose also that $\lim_{x\to 0}f'(x)$ exists. Prove that $f$ is differenciable at zero".

We want to show " $f$ is differenciable at zero", which means we want to show that the limit $\lim_{x\to 0}\frac{f(x) - f(0)}{x-0}$ exists.

Note the numerator is $\lim_{x\to 0}f(x) - f(0) = 0$. This is where we are using the countinuity condition. The denominator is zero as well. Thus, we can use L'Hopital here because the numerator is differenciable. Thus, $\lim_{x\to 0}\frac{f'(x)}{1}$. And this exists by the stated problem.
• May 14th 2008, 07:56 AM
xifentoozlerix
Thank you so much....after sleeping on it I had a similar idea, but you make it seem so simple now (it always seems that way though...simple when you know how).