Induction proof for a function being in $C^{\infty}(\mathbb{R})$ Let $$f(t) = \begin{cases}
     0 &amp; \text{: } x \le 0\\
     e^{-\frac{1}{t^2}}&amp; \text{: } x &gt; 0
\end{cases}$$
Prove that $f(t)$ is in $C^{\infty}(\mathbb{R})$.

I know that for a function to be in $C^{\infty}(\mathbb{R})$ it has to be infinitely times differentiable and each derivative must be continuous. I'm a bit stuck on how to prove this. I'm assuming the answer will need to be proved inductively. Ie:
Suppose the $f^{(n)}(t)$ is continuous. I need to then prove that $f^{n+1}(t)$ exists and is continuous. I'm not quite sure how to do this, I'm new to proofs by induction. I can also image proving continuity at $t=0$ may be a challenge.
Hint.
A standard example of a smooth non-analytic function is
$$
g(x)=\left\{\begin{array}{ll}e^{-\frac{1}{x}} &amp; \text { if } x&gt;0 \\ 0 &amp; \text { if } x \leq 0\end{array}\right.
$$
A detailed proof for smoothness is given in this Wikipedia article:
https://en.wikipedia.org/wiki/Non-analytic_smooth_function#The_function_is_smooth
You can adapt the techniques for your example. A key step is to note that for any positive integer $m$,
$$
\lim_{x\downarrow 0}\frac{e^{-\frac1{x}}}{x^{m}}=0
$$
Alternatively, observe that
$$
f(x)=g(h(x))
$$
where $h(x)=x^2$ for $x\ge 0$ and $h(x)=0$ for $x&lt;0$. You can show that $h$ is smooth. The result then follows from the chain rule.
Find all derivatives of $f(x)$ everywhere on $\mathbb{R}$ except for $t = 0$. Then prove that all of them in fact are continuous at zero.