Let $k \in \nat$ and $\seq{f_n} \subset \cs$ be a Cauchy sequence such that $ \partial^\alpha g_0 = \partial^\alpha \lim_{n \to \infty}f_n = \limv{n}\partial^{\alpha}f_n = g_{\alpha} \quad \forall \alpha: \abs{\alpha} = k $ then $ \partial^\beta g_0 = g_{\beta} \quad \forall \beta: \abs{\beta} = k + 1 $ Since $\partial^0g_0 = g_0$, this applies to all multi-indices. *Proof.* Denote $e_j \in \real^n$ as the vector with a $1$ on the $j$-th position, and $0$s everywhere else. Let $\beta$ be a multi-index such that $\abs{\beta} = k + 1$, then there exists $j \in [n]$ such that $\beta = \alpha + e_j$, and $\partial^\beta f = \frac{\partial}{\partial x_j}\partial^\alpha f$ since the derivatives commute. Since $\partial^\beta f_n \to g_{\beta}$ uniformly, we can interchange the limit with the derivative, $ \frac{\partial}{\partial x_j}g_\alpha = \frac{\partial}{\partial x_j}\limv{n}\partial^\alpha f_n = \limv{n}\frac{\partial}{\partial x_j}\partial^\alpha f_n = \limv{n}\partial^\beta f_n = g_\beta $ By the inductive hypothesis, $g_\alpha = \partial^\alpha g_0$, therefore $ \partial^\beta g_0 = \frac{\partial}{\partial x_j}\partial^\alpha g_0 = \frac{\partial}{\partial x_j}g_\alpha = g_\beta $