$f(x) = \begin{cases} {1-e^{-x^2} \over x^2} \mathrm{, \ if\ } x \neq 0\ \\ 1, \mathrm{if\ } x =0\end{cases} $ Prove $f$ is analytic on $\mathbb{R}$.

Question

I'm studying real-analysis by myself.
In textbook, The exercise problem given as,

$$f(x) = \begin{cases} {1-e^{-x^2} \over x^2} \mathrm{, \ if\ } x \neq 0\ \\ 1, \mathrm{if\ } x =0\end{cases} $$

Exercise : Prove that $f$ is analytic on $\mathbb{R}$.

My proof idea is

1. Show $f(x)$ is infinitely differentiable on $\mathbb{R}$.
2. Find Taylor Series of $f(x)$ and show it converges to $f(x)$ on $\mathbb{R}$.
3. By the definition of analytic functions, we have a result. $\Box$

But I think, There are some problems like,
Q1. Hard to show that $f(x)$ is infinitely differentiable at $x=0$.
Q2. Hard to find the Taylor Series of $f(x)$.

Could you give me some hint or another proof idea? Thank you for your helps. :)

Answer 1

In $\Bbb R\setminus\{0\}$, $f$ is the quotient of two analytic functions, and therefore $f$ is analytic there. On the other hand, near $0$ you have$$f(x)=1-\frac{x^2}2+\frac{x^4}{3!}-\cdots$$and therefore $f$ is also analytic near $0$. So, $f$ is analytic.

Actually, the previous power series converges to $f(x)$ for every real $x$, and the analyticity of $f$ also follows from this.