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  "body": "\n## Proof\n\nWe use the [[sequential characterization of limits|000MXHEY22]] and the limit laws for sequences.\n\n1. Let $H = \\set{h \\ne 0 : a + h \\in I}$. Since $I$ is an open interval containing $a$, there is $r \u003e 0$ with $(a - r, a + r) \\subseteq I$. So $0$ is a [[limit point|07EGCNHAVW]] of $H$.\n2. 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|0F1WR9C83K]], $u'(a) = L$ means $q_u(h) \\to L$ as $h \\to 0$.\n3. Put $s = \\alpha f + \\beta g$. For $h \\in H$, expanding the numerator gives $q_s(h) = \\alpha \\, q_f(h) + \\beta \\, q_g(h)$.\n4. 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 $q_g(h_n) \\to g'(a)$.\n5. The constant sequence $\\alpha, \\alpha, \\dots$ converges to $\\alpha$, since $\\abs{\\alpha - \\alpha} = 0 \u003c \\varepsilon$ for every $\\varepsilon \u003e 0$ (see the [[epsilon-N form|09G6KHS6GY]]). The same holds for $\\beta$.\n6. By the [[limit of a product of sequences|0734RSVF2V]], $\\alpha \\, q_f(h_n) \\to \\alpha f'(a)$ and $\\beta \\, q_g(h_n) \\to \\beta g'(a)$.\n7. By the [[limit of a sum of sequences|0RAM4E07NR]] and step 3, $q_s(h_n) \\to \\alpha f'(a) + \\beta g'(a)$.\n8. This holds for every such sequence $(h_n)$. By the sequential characterization, $q_s(h) \\to \\alpha f'(a) + \\beta g'(a)$ as $h \\to 0$.\n9. So $s = \\alpha f + \\beta g$ is [[differentiable|0A61B60B5R]] at $a$, and $(\\alpha f + \\beta g)'(a) = \\alpha f'(a) + \\beta g'(a)$.\n\n## Idea\n\nThe difference quotient of $\\alpha f + \\beta g$ is the same combination of the difference quotients of $f$ and $g$. Limits respect sums and constant multiples. So the limit of the combination is the combination of the limits.\n",
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