# Math Help - continuity and differentiability

1. ## continuity and differentiability

Hello I would just like you to tell me what would be the best method to demonstrate that the function g is of class $C^1$ on the interval J(I was spoken to apply Taylor Lagrange or Taylor-Young but I did not understand much) I thank:
Here is the statement:

Let f be a function of class $C^2$ over an interval J, such that f (0) = 0.
It is assumed that 0 is an interior point to J. We define the function g on J by:
$g (x) = \frac{f (x)}{(x)}$ if $x \neq 0$
and g (0) = f '(0).

What I have done yet below:

In a first time I apply Taylor Young to see where that leads me

-The formula for Taylor Young in 0 to order 2 of f gives us

$f (x) = f (0) + xf '(0) + x ^ 2 \frac{f"(0)}{2} + o (x ^ 2)$

or f (0) = 0 therefore
$f (x) = xf '(0) + x ^ 2 \frac{f"(0)}{2}$
and
$\ frac{f(x)}{x} = f '(0) + x \frac{f"(0)}{2} = g (x)$

I wonder now if this is fair and whether it will indeed serve me something.
(I just want to say that I'm French and I think you'll find my language can be a little too basic, I hope that this will not prevent you from answering me)

2. The function g is clearly differentiable at all points of J other than 0. In fact, $g'(x) = \frac{xf'(x)-f(x)}{x^2}$ by the usual quotient rule for differentiation.

To see what happens at x=0, you have to find the derivative from first principles: $g'(0) = \lim_{x\to0}\frac{g(x)-g(0)}x$. Using the definitions of g(x) when x≠0 and when x=0, you see that this is equal to $\lim_{x\to0}\frac{f(x)-xf'(0)}{x^2}$. Thus g is differentiable at 0 (with derivative $\tfrac12f''(0)$).

Finally, to show that g is of class $C^1$, you must check that the derivative g' is continuous at 0. To do this, evaluate $\lim_{x\to0}\frac{xf'(x)-f(x)}{x^2}$ (using l'Hôpital's rule) and verify that it is equal to $\tfrac12f''(0)$.