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  "body": "\n## Proof\n\nWe rewrite the conclusion of Taylor's theorem in little o notation.\n\n1. Let $P_k(x) = \\sum_{j=0}^{k} \\frac{f^{(j)}(a)}{j!} (x - a)^j$, the [[Taylor polynomial|0HFSFVMWQB]] of degree $k$ of $f$ at $a$.\n2. As in [[little o notation|04TSHB9H8D]], with $x \\to a$ in place of $x \\to \\infty$, the claim means the following. For each $\\varepsilon \u003e 0$ there is $\\delta \u003e 0$ with $|f(x) - P_k(x)| \\le \\varepsilon |x - a|^k$ for all $x \\in I$ with $|x - a| \u003c \\delta$.\n3. By [[Taylor's theorem|03QHBVW4TT]], there is $h_k \\colon I \\to \\R$ with $h_k(x) \\to 0$ as $x \\to a$ and $f(x) - P_k(x) = h_k(x)(x - a)^k$ for all $x \\in I$.\n4. Let $\\varepsilon \u003e 0$. By the definition of the [[limit|0TXTSEMHR5]] applied to $(-\\varepsilon, \\varepsilon)$, there is $\\delta \u003e 0$ with $|h_k(x)| \u003c \\varepsilon$ for $x \\in I$ and $0 \u003c |x - a| \u003c \\delta$.\n5. For such $x$, $|f(x) - P_k(x)| = |h_k(x)| \\, |x - a|^k \\le \\varepsilon |x - a|^k$.\n6. For $x = a$, step 3 gives $f(a) - P_k(a) = 0$. So the inequality of step 2 holds at $a$ as well.\n7. Hence $f(x) = P_k(x) + o\\big((x - a)^k\\big)$ as $x \\to a$.\n\n## Idea\n\nTaylor's theorem already writes the error as $h_k(x)(x - a)^k$ with $h_k \\to 0$. Saying that the error is $o((x - a)^k)$ is just another way to say that this factor $h_k$ tends to $0$.\n",
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