Differentiability of $f(x) = \exp(-1/x^2), f(0) = 0$ [duplicate]

To answer your new question, if for all $\epsilon > 0$ there exists a $\delta > 0$ such that

$$0 < x < \delta \Rightarrow |f(x) - L| < \epsilon$$

then

$$ 1/x > 1/\delta \Rightarrow |f(x) - L | < \epsilon$$ and

$$t > 1/\delta \Rightarrow |f(1/t) - L| < \epsilon$$

The converse is also true.

That is, $$\lim_{x \to 0^+} f(x) = L \ \Longleftrightarrow \ \lim_{x \to \infty} f(1/x) = L$$

Similarly,

$$\lim_{x \to 0^-} f(x) = L \ \Longleftrightarrow \ \lim_{x \to -\infty} f(1/x) = L$$

Therefore

$$\lim_{x \to 0} f(x) = L \ \Longleftrightarrow \ \lim_{x \to \infty} f(1/x) = \lim_{x \to -\infty} f(1/x) = L$$

As for $\lim_{t\to\infty} te^{-t^2} = \lim_{t\to\infty} \frac{t}{e^{t^2}}$, apply l'Hopital's rule, this limit is equal to $\lim_{t\to\infty} \frac{1}{2te^{t^2}} = 0$.

Similarly, $\lim_{t\to\color{magenta}{-}\infty} te^{-t^2} = 0$ and therefore the derivative at $x = 0$ is zero:

$$\lim_{x\to 0} \frac{f(x) - 0}{x - 0} = \lim_{x\to 0} \frac{e^{-1/x^2}}{x} = 0$$

You should now be able to show that the $n$-th derivative at $x = 0$ is zero for all $n \geq 1$.