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For $n=1$, differentiability at $x_0$ means existence of the finite two-sided limit of a function ${\lim }_{h\to 0}(f(x_0+h)-f(x_0))/h$. More generally, higher-order differentiability at a point is defined recursively: for $n\ge 2$, the derivatives $f^{\prime },\dots ,f^{(n-1)}$ must exist throughout some open interval containing $x_0$, and $f^{(n-1)}$ must be differentiable at $x_0$. With $f^{(0)}=f$, the final condition isThe neighbourhood requirement makes the numerator meaningful for every sufficiently small nonzero $h$. It does not require $f^{(n)}$ to exist at other points of that neighbourhood, or to be continuous at $x_0$.

The zeros of $\cos (\pi /x)$ occur atLet $g(x)=\cos (\pi /x)$. At these nonzero points,The definition of the derivative gives $g(x_k+h)=g^{\prime }(x_k)h+o(|h|)$, henceThe sign function is constant sufficiently near the nonzero point $x_k$, and $(x_k+h{)}^2=x_k^2+O(h)$. As $f(x_k)=0$, multiplication yieldsIts right and left one-sided derivatives are therefore $\pi \mathrm{sign}(x_k)$ and $-\pi \mathrm{sign}(x_k)$ respectively. They differ, so $f$ is not differentiable at any $x_k$. In particular, $x_k\to 0$ as $k\to +\infty$, provingThe absolute value at a simple zero mechanism creates unequal one-sided derivatives. It also shows why the existence of $f^{\prime }(0)$ alone does not imply higher-order differentiability at a point of order two.