Proof of Taylor's theorem by repeated use of L'Hôpital's rule

Proof

We 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 $f^{(k)}(a)$.

1. That $f$ is $k$ times differentiable at $a$ means the following. The higher derivatives $f', \dots, f^{(k-1)}$ 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 $U = I$.
2. For $i \ge 0$ let $q_i(x) = (x - a)^i / i!$. By the power rule, the chain rule and the constant multiple rule, $q_i' = q_{i-1}$ for $i \ge 1$, and $q_0' = 0$.
3. So the $j$-th derivative of $q_i$ is $q_{i-j}$ if $j \le i$, and $0$ if $j > i$. Also $q_0(a) = 1$, and $q_m(a) = 0$ for $m \ge 1$.
4. The Taylor polynomial is $P_k = \sum_{i=0}^{k} f^{(i)}(a) \, q_i$. By step 3, $P_k^{(j)}(a) = f^{(j)}(a)$ for $0 \le j \le k$.
5. 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 $a$.
6. By step 4, $R^{(j)}(a) = 0$ for $0 \le j \le k$.
7. 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 $x \to a$.
8. 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, it tends to $R^{(k)}(a) = 0$.
9. 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 $U$.
10. 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 $x \ne a$.
11. The function $\varphi$ is differentiable at $a$, hence continuous there. So $\varphi(x) \to \varphi(a) = 0$. Also $\psi(x) \to 0$, since $k - j \ge 2$.
12. 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 $x \to a$.
13. By L'Hôpital's rule on $U$ at $a$, $\varphi(x)/\psi(x) \to 0$. This is the claim for $j$.
14. The case $j = 0$ of step 7 says $R(x)/(x - a)^k \to 0$ as $x \to a$.
15. 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 $x \to a$.
16. For $x \ne a$, $f(x) = P_k(x) + R(x) = P_k(x) + h_k(x)(x - a)^k$.
17. 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.