= Solution
{c}
The <function> $f$ is <differentiable> at $c$ if the finite <limit of a function>
$$
f'(c)=\lim_{x\to c,\ x\ne c}\frac{f(x)-f(c)}{x-c}
$$
exists. If it exists, put $L=f'(c)$ and define
$$
\epsilon(x)=\begin{cases}\dfrac{f(x)-f(c)}{x-c}-L,&x\ne c,\\0,&x=c.\end{cases}
$$
The definition of <derivative> gives $\epsilon(x)\to0$ as $x\to c$. For $x\ne c$, rearrangement gives the desired <local linear approximation>
$$
f(x)=f(c)+L(x-c)+\epsilon(x)(x-c).
$$
At $x=c$, this identity holds because both terms containing $x-c$ vanish. Conversely, if such an identity holds, then for $x\ne c$ its <difference quotient> is $L+\epsilon(x)\to L$. Hence $f$ is <differentiable> at $c$, and necessarily $\boxed{L=f'(c)}$.
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