1. ## differentiability

Suppose $\displaystyle g:\mathbb{R}\rightarrow{\mathbb{R}}$ is a twice differentiable function with $\displaystyle g(0)=g'(0)=0$ and $\displaystyle g''(0)=14$. Let $\displaystyle f:\mathbb{R}\rightarrow{\mathbb{R}}$ be defined by
$\displaystyle f(x)=\frac{g(x)}{x}$ if $\displaystyle x\neq{0}$, $\displaystyle 0$ if $\displaystyle x=0$
Prove that $\displaystyle f$ is differentiable at $\displaystyle x=0$ and find $\displaystyle f'(0)$.

Can we use L' Hopital's Rule to solve this question?

2. ## Re: differentiability

Use L' Hopital's Rule to find the limit of f(x) at x=0 first.

f(x) = g(x)/x, since g is a differentiable function,
therefore f(x) must be differentiable somewhere on R.
assume f'(x) exists at point x=k, then

f'(k) = [k g'(k) - g(k) ]/k^2. (by quotient rule)
take above expression become a limit which k approach 0 and apply L' Hopital's Rule to show that f is differentiable at x=0, and hence find f'(0), the answer should be 7.

3. ## Re: differentiability

Originally Posted by piscoau
Use L' Hopital's Rule to find the limit of f(x) at x=0 first.

f(x) = g(x)/x, since g is a differentiable function,
therefore f(x) must be differentiable somewhere on R.
assume f'(x) exists at point x=k, then

f'(k) = [k g'(k) - g(k) ]/k^2. (by quotient rule)
take above expression become a limit which k approach 0 and apply L' Hopital's Rule to show that f is differentiable at x=0, and hence find f'(0), the answer should be 7.
$\displaystyle \lim_{k\to0}f'(k)=\lim_{k\to0}\frac{kg'(k)-g(k)}{k^2}=\lim_{k\to0}\frac{kg''(k)+g'(k)-g'(k)}{2k}=\lim_{k\to0}\frac{g''(k)}{2}=7$

Is this correct?

YES