1. ## Differentiable problem

Suppose that f is continuous on (a,b), if f is also differentiable on (a,b), expect possibly at a single point x0 where lim(x--x0)(f'(x)) exists and is finite. prove that f is diff at x0 and that f' is cont at x0.

can anyone solve this? i have no idea when i processed to lim(f'(x)).. thanks!

2. Originally Posted by shuxue
Suppose that f is continuous on (a,b), if f is also differentiable on (a,b), expect possibly at a single point x0 where lim(x--x0)(f'(x)) exists and is finite. prove that f is diff at x0 and that f' is cont at x0.

can anyone solve this? i have no idea when i processed to lim(f'(x)).. thanks!

$f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}$ ; but since $\lim_{x\to x_0}\left(f(x)-f(x_0)\right)'=\lim_{x\to x_0}f'(x)\,,\,\,\lim_{x\to x_0}(x-x_0)'=\lim_{x\to x_0}1$ both exist and are finite, according to the given info, we can

apply L'Hospital Rule and get $\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}f'(x_0)$ , which exactly means that $f'(x_0)$ exists and also that $f'(x)$ is continuous at $x_0$ (these are definitions!)

Tonio

3. Some (including me) would prefer using directly the mean value theorem. It is really straightforward.

Let us denote $\ell=\lim_{x\to x_0}f'(x)$ (assumed to exist). Let $\epsilon>0$. Choose $\delta>0$ such that $|x-x_0|<\delta$ and $x\neq x_0$ imply $|f'(x)-\ell|<\epsilon$. Then, for all such $x$, the mean value theorem provides $c\in(x,x_0)$ such that $f(x)-f(x_0)=f'(c)(x-x_0)$, hence $\left|\frac{f(x)-f(x_0)}{x-x_0}-\ell\right|=|f'(c)-\ell|< \epsilon$ because $|c-x_0|\leq |x-x_0|<\delta$. This proves that $f$ is differentiable at $x_0$ and $f'(x_0)=\ell$. The assumption $f'(x)\to_{x\to x_0} \ell$ then reads " $f'$ is continuous at $x_0$".