Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 19 / a / ii / Solution

As $x\to 0$ through nonzero values, $1/x^2\to +\infty$ and $e^{-1/x^2}\to 0=f(0)$. Thus $f$ is continuous at zero.

For the derivative at zero use its definition, rather than substituting into a formula obtained away from zero:Indeed, putting $t=1/|h|$ makes the absolute value $t{e}^{-t^2}$, which tends to zero since Gaussian decay dominates every power. Therefore $f$ is continuous and differentiable at zero, with $f^{\prime }(0)=0$. Away from zero, the chain rule gives $f^{\prime }(x)=2x^{-3}{e}^{-1/x^2}$.