The linear-approximation form of the derivative: $f$ is differentiable at $c$ with $f^{\mathrm{\prime }}(c)=L$ if and only if the remainder $r(x)=f(x)-f(c)-L(x-c)$ satisfies ${\lim }_{x\to c}r(x)\mathrm{/}(x-c)=0$; at most one $L$ does so, so $x\mapsto f(c)+L(x-c)$ is the unique affine map approximating $f$ to first order at $c$

Let $A\subseteq \mathbb{R}$, let $f:A\to \mathbb{R}$, let $c\in A$ be a limit point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$) and let $L\in \mathbb{R}$. Write

${\alpha }_L:A\to \mathbb{R},{\alpha }_L(x):=f(c)+L(x-c),$

for the affine map through $(c,f(c))$ of slope $L$, and let $r_L:=f-{\alpha }_L$, that is $r_L(x)=f(x)-f(c)-L(x-c)$.

1. $f$ is differentiable at $c$ with $f^{\mathrm{\prime }}(c)=L$ (The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set) if and only if ${\lim }_{x\to c}\frac{r_L(x)}{x-c}=0,$ the quotient being taken as a function on $A\setminus \{c\}$ (The $\epsilon$-$\delta$ limit ${\lim }_{x\to c}f(x)=L$ of $f:A\to \mathbb{R}$ at a limit point $c$ of $A$).
2. At most one real $L$ satisfies the condition of claim 1. Some real satisfies it exactly when $f$ is differentiable at $c$, and then that real is $f^{\mathrm{\prime }}(c)$.

So among all affine maps through $(c,f(c))$ there is at most one whose error $r_L$ is small compared with $x-c$ near $c$; it exists exactly when $f$ is differentiable at $c$, and its slope is the derivative. This is the sense in which the derivative is a first-order approximation and not merely a quotient.

What the statement does not say. It says nothing about how small $r_L$ is in absolute terms, and nothing about any $x$ away from $c$. The assertion is only that the ratio $r_L(x)\mathrm{/}(x-c)$ tends to $0$; a second-order estimate on $r_L$ needs hypotheses this page does not have.

• Why this is worth stating separately. The quotient form is what one computes with; the remainder form is what generalises, since it never divides by the increment and so survives verbatim in settings where the increment is not a number one may divide by. Nothing on this page needs that generality, but the equivalence is what licenses the phrase "best linear approximation" used informally elsewhere, and the phrase is otherwise unearned.

• The two forms are the same $\epsilon$-$\delta$ condition, not two conditions that happen to agree. Step 1.1 is an identity of functions on $A\setminus \{c\}$, and everything after it is bookkeeping. In particular the proof spends no limit theorem at all: no algebra of limits, no sequences and no choice principle.

• Uniqueness is the whole of claim 2, and it is inherited. It comes from At a limit point of the domain a function has at most one limit, the same lemma that lets $f^{\mathrm{\prime }}(c)$ be written at all (The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set). Without a limit point of the domain there is no uniqueness anywhere in sight, and the phrase "the best approximation" would name nothing.