Proof of the linearity of differentiation

Proof

We use the sequential characterization of limits and the limit laws for sequences.

1. Let $H = \set{h \ne 0 : a + h \in I}$. Since $I$ is an open interval containing $a$, there is $r > 0$ with $(a - r, a + r) \subseteq I$. So $0$ is a limit point of $H$.
2. For a function $u \colon I \to \R$ and $h \in H$, write $q_u(h) = \frac{u(a + h) - u(a)}{h}$. By the definition of the derivative, $u'(a) = L$ means $q_u(h) \to L$ as $h \to 0$.
3. Put $s = \alpha f + \beta g$. For $h \in H$, expanding the numerator gives $q_s(h) = \alpha \, q_f(h) + \beta \, q_g(h)$.
4. Let $(h_n)$ be a sequence in $H$ with $h_n \to 0$. By the sequential characterization, $q_f(h_n) \to f'(a)$ and $q_g(h_n) \to g'(a)$.
5. The constant sequence $\alpha, \alpha, \dots$ converges to $\alpha$, since $\abs{\alpha - \alpha} = 0 < \varepsilon$ for every $\varepsilon > 0$ (see the epsilon-N form). The same holds for $\beta$.
6. By the limit of a product of sequences, $\alpha \, q_f(h_n) \to \alpha f'(a)$ and $\beta \, q_g(h_n) \to \beta g'(a)$.
7. By the limit of a sum of sequences and step 3, $q_s(h_n) \to \alpha f'(a) + \beta g'(a)$.
8. This holds for every such sequence $(h_n)$. By the sequential characterization, $q_s(h) \to \alpha f'(a) + \beta g'(a)$ as $h \to 0$.
9. So $s = \alpha f + \beta g$ is differentiable at $a$, and $(\alpha f + \beta g)'(a) = \alpha f'(a) + \beta g'(a)$.