How to prove $\lim_{x\to 0}\sum^\infty_{k=0}\frac{(-1)^k}{k!}x^{-2k}=0$? How to prove $$\lim_{x\to 0}\sum^\infty_{k=0}\frac{(-1)^k}{k!}x^{-2k}=0$$ as if you do not recognize it as $e^{-1/x^2}$?

My attempt:
I can come up with an approach, which poorly pretends that I know nothing about $e^{-1/x^2}$.
Let $$f(x)=\sum^\infty_{n=0}\frac{x^{-2n}}{n!}$$
Then, $$f(x) \cdot \sum^\infty_{k=0}\frac{(-1)^k}{k!}x^{-2k} = \sum^\infty_{n=0}\sum^\infty_{k=0}\frac{(-1)^k}{n!k!}x^{-2(n+k)}$$
The coefficient of $x^{-2j}$ ($j\in\mathbb N$) on the right hand side is
$$[x^{-2j}]=\sum^j_{n=0}\frac{(-1)^n}{n!(j-n)!}=\frac1{j!}\sum^j_{n=0}\binom{j}{n}(-1)^n(1)^{j-n}=\frac1{j!}(-1+1)^j=0$$
(Note that binomial theorem is utilized.)
Also, $[x^0]=1$. Hence,
$$\sum^\infty_{k=0}\frac{(-1)^k}{k!}x^{-2k} =\frac1{f(x)}$$
We then have $$\lim_{x\to 0}f(x)>\lim_{x\to 0}\frac1{x^2}=+\infty$$
$$\lim_{x\to 0}\frac1{f(x)}=0$$
and so does the series.
Clearly I know something about $e^{-1/x^2}$, otherwise I would not have been able to define such a nice $f(x)$ that multiplies with the series to give $1$.

The simplest answer will be accepted.
NB: I am looking for alternative methods.
 A: The following proof proceeds by studying the differential equation associated to the function $f$. I tried my best so that the proof employs the standard ideas in the theory of ordinary differential equation.


*

*By the substitution $z = x^{-2}$, it suffices to consider the function
$$ g(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} z^n $$
and then investigate the limit of $g(z)$ as $z\to+\infty$. To this end, notice that $g$ satisfies the differential equation
$$ g'(z) = -g(z). \tag{DE}$$

*Now assume that $g \geq 0$ on $[a, b]$ for some $a < b$. By the Mean Value Theorem,
$$ g(a) - g(b) \stackrel{\exists \xi \in (a,b)}{=} -g'(\xi)(b-a) \stackrel{\text{(DE)}}{=} g(\xi)(b-a). $$
Also, $\text{(DE)}$ shows that $g$ is non-increasing on $[a, b]$, and so, $g(a) \geq g(\xi) \geq g(b)$ holds. Plugging the above equality to this and rearranging, we get the following implication
$$ g \geq 0 \text{ on } [a, b] \quad \Rightarrow \quad (1-(b-a))g(a) \leq g(b) \leq \frac{g(a)}{1+(b-a)}. \tag{1}$$
In particular, if $b-a < 1$ and $g(b) = 0$, then $\text{(1)}$ shows that $g(a) = 0$ as well, hence $g \equiv 0$ on $[a, b]$.

*We show that $g > 0$ on all of $[0, \infty)$. Assume otherwise. Since $g(0) = 1$, this implies that $g$ has a positive zero. Since $g$ is continuous, we can find the smallest positive zero of $g$, which we call $\alpha$. Since $g \geq 0$ on $[0, \alpha]$, $\text{(1)}$ allows to find a positive zero of $g$ smaller than $\alpha$, a contradiction. This shows that the claim is true.

*By $\text{(1)}$ applied to $[0, x]$, we have $ 0 < g(x) \leq \frac{g(0)}{1+x} $. Then by the squeezing theorem, $g(x) \to 0$ as $x\to\infty$, and we are done.
