> [!NOTE] Lemma (aka Local linearisation) > Let $I$ be an [[Open Real Interval|open interval]], $f:I\to \mathbb{R}$ and $c\in I$. Then $f$ is [[Fréchet Differentiation|differentiable]] at $c$ iff there exists $A\in\mathbb{R}$ and a function $\varepsilon$ with the properties that for all $x$ $f(x)-f(c)=A(x-c)+\varepsilon(x)(x-c),$$\varepsilon(c)=0$ and $\varepsilon$ is [[Continuous Real Function|continuous]] at $c$: $\varepsilon(x)\to {0}$ as $x\to c$. If this happens $A=f'(c)$. > ###### Proof Suppose the condition holds then for $x\neq c,$ $\frac{f(x)-f(c)}{x-c} = A+\varepsilon(x)$and this approaches $A$ as $x\to c$. Hence $f$ is differentiable with derivative $f'(c)=A$. Now suppose $f$ is differentiable. Set $A=f'(c)$ and defined $\varepsilon$ as follows $\varepsilon(x)=\begin{cases} \frac{f(x)-f(c)}{x-c} - A & \text{if } x \neq c \\ 0 & \text{if } x=c \end{cases}$If $x\neq c$ then $f(x)-f(c)=A(x-c)+\varepsilon(x)(x-c)$ holds because of the way $\varepsilon$ is defined, while if $x=c$ the formula is obvious. To check that $\varepsilon(x)\to{0}$ as $x\to c$, observe that $\lim_{ x \to c } \varepsilon(x) = \lim_{ x \to c } \left( \frac{f(x)-f(c)}{x-c} - f'(c) \right) = 0 $ **Proof**: Follows directly from [[Fréchet Differentiation]].