Let define $f:\mathbb{R}\rightarrow\mathbb{R}$ such as
$f(x)=\frac{\sin(\sqrt{x})}{\sqrt{x}}$ for $x>0$ and $f(x)=1$ for $x=0$
Prove that $f$ has right-sided all-order derivatives in $0$.
My approach:
I find out that $f$ can be written as the power series $$ f(x)=\sum_{n=0}^{\infty}\frac{(-1)^k}{(2k+1)!}x^k $$ which has an infinite radius of convergence so the $n$-th derivative of $f$ in zero would be $$ f^{(n)}(0)=\frac{(-1)^n n!}{(2n+1)!} $$ but I do not know how to connect these facts with the right-sided derivative of this function
Since$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots,$$you have$$\frac{\sin\left(\sqrt x\right)}{\sqrt x}=1-\frac x{3!}+\frac{x^2}{5!}-\frac{x^3}{7!}+\cdots$$when $x\geqslant0$, from which it follows that, indeed, $f$ has right-sided derivatives of all orders at $0$.