# Statement Let $U \subset \mathbb{R}^{n}$ be [[Open subsets of metric spaces|open]]. If $f:U \to \mathbb{R}^{m}$ is [[Fréchet Differentiation|differentiable]] at $p\in U$, then $f$ is [[Continuous maps|continuous]] at $p$. # Proof Let $\varepsilon>0$. Since $f$ is differentiable at $p$, there exists $\delta_{1}>0$ such that $0<\lVert h \rVert<\delta_{1}$ implies $\lVert f(p+h) - f(p) - Df(p)(h) \rVert <\varepsilon \lVert h \rVert .$The triangle inequality yields $\begin{align} \lVert f(p+h) -f(p) \rVert &\leq \lVert f(p+h) - f(p) - Df(p)(h) \rVert + \lVert Df(p)(h) \rVert \\ & < (\varepsilon + \lVert Df(p) \rVert_{op} ) \lVert h \rVert < \varepsilon \end{align}$if $\lVert h \rVert< \min(\delta_{1}, \varepsilon/(\varepsilon + \lVert Df(p) \rVert_{op}))$, where $\lVert \cdot \rVert_{\text{op}}$ denotes the [[Operator Norm|operator norm]] (one may even use the [[Frobenius Norm|Frobenius norm]] instead). This completes the proof.