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!

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

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

Tonio

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

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