Prove $h(x)$ is analytic $\forall x \in \mathbb R $. Given the function
$$
g(x) =
\begin{cases}
\frac{\sin x}{x},  & \text{if $x\neq0$ } \\[2ex]
1, & \text{if $x=0$}
\end{cases}
$$
How can I prove the function $h(x)=g(\sqrt|x|)$ has a power series development at $x=0$ and find its radius of convergence? Furthermore, how can I prove it is analytic $\forall x \in \mathbb R $? Could someone give me a hint to start the proof. Thanks in advance.
 A: The given function $h$ is not differentiable at $x=0$, and therefore certainly cannot be analytic at $x=0$.
To see this, we expand the definition of differentiability.  Namely, in order for $h$ to be differentiable at $x=0$, the following limit would have to exist:
$$\lim_{t\to 0} \frac{h(t) - h(0)}{t} = \lim_{t\to 0} \frac{\frac{\sin(\sqrt{|t|})}{\sqrt{|t|}} - 1}{t} = \lim_{t\to 0} \frac{\sin(\sqrt{|t|}) - \sqrt{|t|}}{t \sqrt{|t|}}.$$
Now considering the right-hand limit, if $t > 0$ then the function in the limit reduces to
$$\lim_{t\to 0^+} \frac{\sin(\sqrt{t}) - \sqrt{t}}{t\sqrt{t}}.$$
From here, observe that $\lim_{u\to 0} \frac{\sin(u) - u}{u^3} = -\frac{1}{6}$.  (Either use the Taylor series expansion of $\sin$, or apply l'Hopital's rule three times.)  Therefore, since $\sqrt{t} \to 0$ as $t\to 0^+$, we can conclude that
$$\lim_{t\to 0^+} \frac{\sin(\sqrt{t}) - \sqrt{t}}{t\sqrt{t}} = -\frac{1}{6}.$$
A similar calculation will show that
$$\lim_{t\to 0^-} \frac{\sin(\sqrt{|t|}) - \sqrt{|t|}}{t \sqrt{|t|}} = \lim_{t\to 0^-} \frac{\sin(\sqrt{-t}) - \sqrt{-t}}{t \sqrt{-t}} = \frac{1}{6}.$$
Since the right-hand and left-hand limits are not equal, the limit in the definition of differentiability does not exist; so $h$ is not differentiable at $x=0$.
