Show that $f^{[n]}(0)=0$ for all $n=0,1,2...$

Question

Let $$f(x)=\left\{ {\matrix{ {{e^{ - {1 \over {{x^2}}}}},x \ne 0} \cr {0,x = 0} \cr } } \right.$$

Show that $f^{[n]}(0)=0$ for all $n=0,1,2,\cdots$

The proof:
First note that for $x\ne 0$: $f'(x)=\frac{2}{x^3} e^{-\frac{1}{x^2}}$

Consider the polynomial function $P_1(x) = 2x^3$, then we have:

$$f'(x) = P_1\left(\frac{1}{x}\right)e^{-\frac{1}{x^2}}$$

Since,

$$\lim\limits_{x\to 0} \frac{1}{x^n} e^{-\frac{1}{x^2}} = 0 $$

for any integer $n\ge 0$:

$$\lim\limits_{x\to 0} P\left(\frac{1}{x}\right) e^{-\frac{1}{x^2}} = 0 $$

for any polynomial $P(x)$.

Then, by induction we prove there exists $P_n(x)$ such that: $f^{(n)}(x) = P_n\left(\frac{1}{x}\right)e^{-\frac{1}{x^2}}$

My Question:
I don't understand the base case: Why are the two limits approaches $0$? (And where the $\frac{1}{x^n}$ came from?). It looks like black-magic :)

Thanks in advance.

Answer 1

To make sure consider $u=\frac{1}{x}$. Then 1st limit $$\lim\limits_{u\to \infty} \frac{e^{-u^2}}{u^{n}} \quad \forall n \in \mathbb{N}$$ Using many well-known methods ($\varepsilon-\delta$, for example) you can prove that it is $0$. For better understanding it means that exponential decays faster than any power function. The second is a corollary from the 1st: it's no difference between polynomial or its greater power summand in denominator