Curious fact about the exponential function: $\lim_{n\rightarrow +\infty} \frac{\partial^n e^x}{\partial x^n} \overset{?}{\neq} 0$

Question

I'm really interested in the concept of infinity in mathematics, and I would like to ask a question about a curious fact:

The function $f(x) = e^x$, which we all know that all its derivates are are equal to $e^x$, can be written as a Taylor serie:

$$\displaystyle e^x = \sum_{k=0}^\infty \frac{x^k}{k!} $$

It is thus, at first sight, expressed as a polynom $\mathcal{P}_n(x)$ with $n \rightarrow +\infty$, with:

$$\mathcal{P}_n(x) \equiv a_0 + a_1x + a_2x^2 + \dots + a_n x^n \ \ \ (a_i \in \mathbb{R} \ \forall i \in \mathbb{N}_, \ a_n \neq 0)$$ Let's consider that polynom $\mathcal{P}_n(x)$ as a function $p_n(x) \in C^\infty(\mathbb{R}, \mathbb{R})$. We know that:

$$\frac{\partial^n p_n(x)}{\partial x^{n}} \in \mathbb{R}_0 \Longrightarrow { \frac{\partial^{n+1} p_n(x)}{\partial x^{n+1}} = 0 \ \ \mathrm{as \ long \ as} \ n \in \mathbb{R}}$$

Now, let's go back to the complete Taylor development of $e^x$, which I'll write as $\mathcal{P}_{m}(x)$. Can we really say that $m \rightarrow \infty$ ? Actually, not really ! Indeed:

$$\lim_{k\rightarrow +\infty} \frac{x^k}{k!} = 0$$

(Notice that is the equation above was false, the Taylor serie wouldn't converge). I deduce, then, that:

$$\lim_{n\rightarrow +\infty} \frac{\partial^n \mathcal{P}_m(x)}{\partial x^n} = 0$$

$$\Longleftrightarrow \lim_{n\rightarrow +\infty} \frac{\partial^n e^x}{\partial x^n} = \color{red}{0 \neq e^x}$$

I'm not a mathematician, and I believe I just played unconsciously with the infinite.

I'm particularly interested in those calculations because I wonder if there is a way to "classify" the different infinites (for instance, express that $\displaystyle \lim_{x \rightarrow + \infty} e^x \neq \lim_{x \rightarrow + \infty} x$, even if they are both equal to $+\infty$).

To conclude:

1. Is my development wrong ? Where and why ?
2. Does this "classification" of infinites already exist ? Where could I learn more about it ?

Answer 1

Your explanation is wrong. I will start with the correct parts.

• Your definition of $e^{x}$, and its characterisation as a limit of polynomials, is fine.

$$\forall x, e^{x} = \lim_{n \to \infty}P_{n}(x) = \lim_{n \to \infty}\sum_{i=0}^{n}\frac{x^{i}}{i!}$$

• All polynomials are eventually annihilated by repeated differentiation. In particular, for any polynomial $P$ the sequence of derivates tends to zero. (since it eventually hits 0 and stays there).

$$\forall P, \forall x, \lim_{m \to \infty}\frac{\mathrm{d}^{m}P }{\mathrm{d}x^{m}}(x) \to 0$$

Now for the incorrect parts.

• You say "let $P_m(x)$ be the complete Taylor development of $e^{x}$". If it is the complete development, then what is $m$? I assume you actually mean "let $P_m(x)$ be the partial sums defined above", which is a polynomial of degree $m$. You imply that there is contradiction between $P_{m}(x) \to e^{x}$ as $m \to \infty$ and $x^k/k! \to 0$ as $k \to \infty$. There is no such contradiction.

Finally, the key issue: You state correctly that

$$\lim_{n \to \infty}\frac{\mathrm{d}^{n}P_{m}(x) }{\mathrm{d}x^{n}}(x) \to 0.$$

You then take the limit as $m \to \infty$ and get the correct statement

$$\lim_{m \to \infty} \lim_{n \to \infty}\frac{\mathrm{d}^{n}P_{m}(x) }{\mathrm{d}x^{n}}(x) \to 0.$$

What you cannot do is swap those limits. This is your key mistake. The real limit is the obvious one.

$$\lim_{n \to \infty}\lim_{m \to \infty} \frac{\mathrm{d}^{n}P_{m}(x) }{\mathrm{d}x^{n}}(x) \to e^{x}.$$