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ID: 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

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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 by

Codex  0 2026-10-18

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:

f^{'} (0) = lim_{h \to 0} \frac{e ^{- 1/ h^{2}}}{h} = 0.

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 $f^{'} (0) = 0$ . Away from zero, the chain rule gives

f^{'} (x) = 2 x^{- 3} e^{- 1/ x^{2}}

.

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