With $f \in C^2([0,1])$ prove that $h(x)=\frac{f(x)}{x^{1/2}}- \frac{f(0)}{x^{1/2}}- x^{1/2}f'(0)$ is in $C^1([0,1])$

Question

Question:

If $f \in C^2([0,1])$, then prove that $h(x)=\frac{f(x)}{x^{1/2}}- \frac{f(0)}{x^{1/2}}- x^{1/2}f'(0)$ is in $C^1([0,1])$

Answer:

My best attempt from now.

1-
I have tried different ways below one of my attempt that is not conclusive.
$h'(x) = \frac{f'(x)}{x^{1/2}}-\frac{f(x)}{2x^{3/2}}+\frac{f(0)}{2x^{3/2}}-\frac{f'(0)}{2x^{1/2}}=\frac{2xf'(x)-xf'(0)-f(x)-f(0)}{2x^{3/2}}$
So the only difficult point for the continuity is at $x=0$.

2-
Let us write $f(x)=f(0)+f'(0)x+O(x^2)$ and $f'(x)=f'(0)+O(x)$. Now replace this in the formula of $h'(x)$. We get:
$h'(x)=\frac{2xf'(0)+2xO(x)-xf'(0)-f(0)-f'(0)x-O(x^2)-f(0)}{2x^{3/2}}=\frac{-2f(0)+O(x^2)}{2x^{3/2}}=\frac{O(x^2)}{x^{3/2}}=O(x^{1/2})$

Is this correct? (mainly the part "2-").
Indeed I feel a weakness in my prove as I ve written: $\frac{-2f(0)+O(x^2)}{2x^{3/2}}=\frac{O(x^2)}{x^{3/2}}$ and i am not sure it this is true as $\frac{-2f(0)+O(x^2)}{2x^{3/2}}= \frac{-2f(0)}{2x^{3/2}}+\frac{O(x^2)}{2x^{3/2}}$ and $\frac{-2f(0)}{2x^{3/2}}$ goes to $+/- \infty$ when $x \rightarrow 0$ .

I've tried too using L'hopital's rule but it doesn't work. Any help will be appreciated.

EDIT: As I made a calculation error I've edited my post. I will be happy to know if it is correct.
More over I am not totally sure on how work a demonstration of the continuity using big $O$ notation just like the one I've just written. Thus I will be happy to have a more/rigorous explanation about it.

Answer 1

After the review of an other person (that I greatly thank) I ve made an error at the numerator where it is $+f(0)$. And in fact this is what we get.

1-
I have tried different ways below one of my attempt that is not conclusive.
$h'(x) = \frac{f'(x)}{x^{1/2}}-\frac{f(x)}{2x^{3/2}}+\frac{f(0)}{2x^{3/2}}-\frac{f'(0)}{2x^{1/2}}=\frac{2xf'(x)-xf'(0)-f(x)-f(0)}{2x^{3/2}}$
So the only difficult point for the continuity is at $x=0$.

2-
Let writte $f(x)=f(0)+f'(0)x+O(x^2)$ and $f'(x)=f'(0)+O(x)$ now replace this in the formula of $h(x')$. We get:
$h'(x)=\frac{2xf'(0)+2xO(x)-xf'(0)-f(0)-f'(0)x-O(x^2)+f(0)}{2x^{3/2}}=\frac{O(x^2)}{2x^{3/2}}=\frac{O(x^2)}{x^{3/2}}=O(x^{1/2})$

In such a case my answer is correct.