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  "body": "\n## Proof\n\nWe show that $(f(x) - P_k(x))/(x - a)^k \\to 0$, using L'Hôpital's rule $k - 1$ times and then the definition of $f^{(k)}(a)$.\n\n1. That $f$ is $k$ times differentiable at $a$ means the following. The [[higher derivatives|084NGSTWM7]] $f', \\dots, f^{(k-1)}$ exist on an open interval $U \\subseteq I$ containing $a$, and $f^{(k-1)}$ is differentiable at $a$. For $k = 1$ this says only that $f$ is differentiable at $a$, and we take $U = I$.\n2. For $i \\ge 0$ let $q_i(x) = (x - a)^i / i!$. By the [[power rule|0JZGW75C5Z]], the [[chain rule|0RXX2PT1MQ]] and the [[constant multiple rule|0Z9JB71BS0]], $q_i' = q_{i-1}$ for $i \\ge 1$, and $q_0' = 0$.\n3. So the $j$-th derivative of $q_i$ is $q_{i-j}$ if $j \\le i$, and $0$ if $j \u003e i$. Also $q_0(a) = 1$, and $q_m(a) = 0$ for $m \\ge 1$.\n4. The [[Taylor polynomial|0HFSFVMWQB]] is $P_k = \\sum_{i=0}^{k} f^{(i)}(a) \\, q_i$. By step 3, $P_k^{(j)}(a) = f^{(j)}(a)$ for $0 \\le j \\le k$.\n5. Let $R(x) = f(x) - P_k(x)$ for $x \\in I$. By step 1, $R^{(j)}$ exists on $U$ for $j \\le k - 1$, and $R^{(k-1)}$ is differentiable at $a$.\n6. By step 4, $R^{(j)}(a) = 0$ for $0 \\le j \\le k$.\n7. We prove by backward induction on $j$, from $j = k - 1$ down to $j = 0$, that $R^{(j)}(x) / (x - a)^{k - j} \\to 0$ as $x \\to a$.\n8. Case $j = k - 1$. The quotient is $(R^{(k-1)}(x) - R^{(k-1)}(a))/(x - a)$, since $R^{(k-1)}(a) = 0$. By the definition of the [[derivative|0F1WR9C83K]], it tends to $R^{(k)}(a) = 0$.\n9. Step. Let $0 \\le j \\le k - 2$, and assume the claim for $j + 1$. Put $\\varphi = R^{(j)}$ and $\\psi(x) = (x - a)^{k - j}$ on $U$.\n10. Both $\\varphi$ and $\\psi$ are differentiable on $U \\setminus \\set{a}$, since $j + 1 \\le k - 1$. Also $\\psi'(x) = (k - j)(x - a)^{k-j-1} \\ne 0$ for $x \\ne a$.\n11. The function $\\varphi$ is differentiable at $a$, hence [[continuous|07ZMX4ANJS]] there. So $\\varphi(x) \\to \\varphi(a) = 0$. Also $\\psi(x) \\to 0$, since $k - j \\ge 2$.\n12. By the induction hypothesis, $\\varphi'(x)/\\psi'(x) = \\dfrac{1}{k - j} \\cdot \\dfrac{R^{(j+1)}(x)}{(x - a)^{k - j - 1}} \\to 0$ as $x \\to a$.\n13. By [[L'Hôpital's rule|0JWT4PPQ7B]] on $U$ at $a$, $\\varphi(x)/\\psi(x) \\to 0$. This is the claim for $j$.\n14. The case $j = 0$ of step 7 says $R(x)/(x - a)^k \\to 0$ as $x \\to a$.\n15. Define $h_k \\colon I \\to \\R$ by $h_k(x) = R(x)/(x - a)^k$ for $x \\ne a$, and $h_k(a) = 0$. By step 14, $h_k(x) \\to 0$ as $x \\to a$.\n16. For $x \\ne a$, $f(x) = P_k(x) + R(x) = P_k(x) + h_k(x)(x - a)^k$.\n17. For $x = a$, $f(a) = P_k(a)$ by step 4 with $j = 0$. Also $(a - a)^k = 0$, since $k \\ge 1$. So the formula holds at $a$ too.\n\n## Idea\n\nThe remainder $R = f - P_k$ and its first $k$ derivatives all vanish at $a$, because $P_k$ is built to match them. Each use of L'Hôpital's rule removes one power of $x - a$ and one derivative of $R$. After $k - 1$ steps we reach a difference quotient of $R^{(k-1)}$, whose limit is $R^{(k)}(a) = 0$.\n",
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