严谨求解分段函数在0点的n阶导数（分析学）

Alright, let's work through this problem carefully—since you're diving into Baby Rudin's differentiation section, we'll use the rigorous, induction-based approach that aligns with the book's style, leaning on the limit result you already have.

First, let's restate the function clearly:
$$f(x) = \begin{cases} e^{-1/x} , & x > 0 \ 0, &x \leq 0 \end{cases}$$

We know from the problem that for every $n \in \mathbb{N}$, $\displaystyle \lim_{x\to 0^+} \frac{e^{-1/x}}{xn} = 0$. We'll use this to prove $f^{(n)}(0) = 0$ for all natural numbers $n$ via mathematical induction.

Step 1: Base Case ($n=1$, First Derivative at 0) #

To find $f'(0)$, we use the definition of the derivative and check left and right limits separately:

• Left Derivative ($h \to 0^-$): For $h < 0$, $f(h) = 0$, so:
  $$\lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{0 - 0}{h} = 0$$
• Right Derivative ($h \to 0^+$): For $h > 0$, $f(h) = e^{-1/h}$, so:
  $$\lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{e^{-1/h}}{h}$$
  This is exactly the limit you mentioned with $n=1$, which equals 0.

Since left and right derivatives are equal, $f'(0) = 0$. Our base case holds.

Step 2: Inductive Step #

Suppose that for some $k \in \mathbb{N}$, $f^{(k)}(0) = 0$. We need to show $f^{(k+1)}(0) = 0$.

First, note a key observation about $f^{(k)}(x)$ when $x > 0$: every derivative of $e^{-1/x}$ for $x>0$ takes the form $e^{-1/x} \cdot P_k(1/x)$, where $P_k$ is a polynomial in $1/x$. You can verify this quickly with induction too:

• For $k=1$, $f'(x) = e^{-1/x} \cdot \frac{1}{x^2}$ (polynomial $P_1(t) = t^2$ where $t=1/x$)
• If $f^{(k)}(x) = e^{-1/x} \cdot P_k(1/x)$, then differentiating gives $f^{(k+1)}(x) = e^{-1/x} \cdot \left( \frac{1}{x^2} P_k(1/x) - \frac{1}{x^2} P_k'(1/x) \right)$, which is $e^{-1/x}$ times another polynomial in $1/x$.

Now, compute $f^{(k+1)}(0)$ using the derivative definition:
$$f^{(k+1)}(0) = \lim_{h \to 0} \frac{f^{(k)}(h) - f^{(k)}(0)}{h} = \lim_{h \to 0} \frac{f^{(k)}(h)}{h}$$

Again, split into left and right limits:

• Left Limit ($h \to 0^-$): For $h < 0$, $f(h) = 0$, so all its derivatives are also 0. Thus $f^{(k)}(h) = 0$, so:
  $$\lim_{h \to 0^-} \frac{0}{h} = 0$$
• Right Limit ($h \to 0^+$): Here, $f^{(k)}(h) = e^{-1/h} \cdot P_k(1/h)$. Let $t = 1/h$, so as $h \to 0^+$, $t \to +\infty$. Rewrite the limit:
  $$\lim_{h \to 0^+} \frac{e^{-1/h} \cdot P_k(1/h)}{h} = \lim_{t \to +\infty} e^{-t} \cdot P_k(t) \cdot t$$
  Since $P_k(t)$ is a polynomial, $P_k(t) \cdot t$ is another polynomial (say $Q(t)$) of some degree $m$. We know from your given limit (rewritten for $t$) that $\lim_{t \to +\infty} t^m e^{-t} = 0$ for any $m \in \mathbb{N}$. So this limit equals 0.

Left and right limits are equal, so $f^{(k+1)}(0) = 0$. The inductive step holds.

Conclusion #

By mathematical induction, $f^{(n)}(0) = 0$ for all $n \in \mathbb{N}$.

内容的提问来源于stack exchange，提问作者Dominated Convergence Theorem