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  "body": "\n## Proof\n\nWe unfold the definition of the limit in both directions.\n\n1. Let $V$, $W$, $U$, $f$, $x$ and $A$ be as in the statement, with $V$ and $W$ [[normed vector spaces|00A0RPAQ8A]]. For $h \\in V$ with $x + h \\in U$, put $r(h) = f(x + h) - f(x) - A h$.\n2. Then $f(x + h) = f(x) + A h + r(h)$ for each such $h$, and $r(0) = f(x) - f(x) - A 0 = 0$.\n3. By the definition of the [[Fréchet derivative|08EVNYZF42]], $A$ is the derivative of $f$ at $x$ exactly when $\\|r(h)\\| / \\|h\\| \\to 0$ as $h \\to 0$. The limit is over $h \\ne 0$ with $x + h \\in U$.\n4. Suppose first that $A$ is the derivative. Let $\\varepsilon \u003e 0$. By step 3 there is $\\delta \u003e 0$ with $\\|r(h)\\| / \\|h\\| \u003c \\varepsilon$ for $0 \u003c \\|h\\| \u003c \\delta$ and $x + h \\in U$.\n5. For these $h$, multiplying by $\\|h\\| \u003e 0$ gives $\\|r(h)\\| \\le \\varepsilon \\|h\\|$.\n6. For $h = 0$, step 2 gives $\\|r(0)\\| = 0 = \\varepsilon \\|0\\|$. So $\\|r(h)\\| \\le \\varepsilon \\|h\\|$ for each $h$ with $\\|h\\| \u003c \\delta$ and $x + h \\in U$. Thus $r(h) = o(\\|h\\|)$.\n7. Conversely, suppose that $r(h) = o(\\|h\\|)$. Let $\\varepsilon \u003e 0$. Apply the little-o condition to $\\varepsilon / 2$. This gives $\\delta \u003e 0$ with $\\|r(h)\\| \\le (\\varepsilon / 2) \\|h\\|$ for $\\|h\\| \u003c \\delta$ and $x + h \\in U$.\n8. For $0 \u003c \\|h\\| \u003c \\delta$ with $x + h \\in U$, divide by $\\|h\\|$. This gives $\\|r(h)\\| / \\|h\\| \\le \\varepsilon / 2 \u003c \\varepsilon$.\n9. So $\\|r(h)\\| / \\|h\\| \\to 0$ as $h \\to 0$. By step 3, $A$ is the Fréchet derivative of $f$ at $x$.\n\n## Idea\n\nThe two conditions say the same thing about the error $r(h)$. The limit form divides by $\\|h\\|$, and the little-o form multiplies back. The only extra point is $h = 0$, where the error is $0$, so the inequality holds there too.\n",
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