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  "body": "\n## Proof\n\n1. Since $f$ is [[differentiable|0A61B60B5R]] at $a$, it has a [[derivative|0F1WR9C83K]] $L = f'(a)$. For $h \\ne 0$ with $a + h \\in I$, write $q(h) = \\frac{f(a + h) - f(a)}{h}$. Then $q(h) \\to L$ as $h \\to 0$.\n2. By the [[epsilon-delta form of limits|0GBZDAFPX8]] with $\\varepsilon = 1$, there is $\\delta_1 \u003e 0$ with $\\abs{q(h) - L} \u003c 1$ whenever $0 \u003c \\abs{h} \u003c \\delta_1$ and $a + h \\in I$.\n3. For such $h$, the [[triangle inequality|03HF5PMY5T]] gives $\\abs{q(h)} \\le \\abs{q(h) - L} + \\abs{L} \u003c \\abs{L} + 1$.\n4. So $\\abs{f(a + h) - f(a)} = \\abs{q(h)} \\, \\abs{h} \\le (\\abs{L} + 1) \\abs{h}$ for these $h$. This also holds for $h = 0$, where both sides are $0$.\n5. Let $\\varepsilon \u003e 0$. Put $\\delta = \\min\\left(\\delta_1, \\frac{\\varepsilon}{\\abs{L} + 1}\\right) \u003e 0$.\n6. Let $x \\in I$ with $\\abs{x - a} \u003c \\delta$, and put $h = x - a$. By step 4, $\\abs{f(x) - f(a)} \\le (\\abs{L} + 1) \\abs{h} \u003c (\\abs{L} + 1) \\cdot \\frac{\\varepsilon}{\\abs{L} + 1} = \\varepsilon$.\n7. By the [[epsilon-delta form of continuity|0W8YHHGCJG]], $f$ is [[continuous at|0EEE2CZB2A]] $a$.\n\n## Idea\n\nNear $a$, the difference quotient stays close to $f'(a)$, so it is bounded. Then $f(x) - f(a)$ is a bounded number times $x - a$. This product is small when $x$ is close to $a$.\n",
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