# “Proof” that if f(x) ~x, e^f(x) ~ e^x

While it is not true that $f(x)\sim x \implies e^{f(x)}\sim e^x,$ I can't spot the error in this "proof" by induction--or at least I can't articulate it well.

Let $f(x)\sim x$ and $x > 1$

P(1): $1 \sim 1, f(x) \sim x,$ and $\lim_{x \to \infty} \frac{1+f(x)}{1+x} = \frac{1}{1+x}+\frac{f(x)}{1+x} = 0 +1 = 1.$

Assume P(k): $$~\lim_{x \to \infty} \frac{\sum_0^{k-1}f(x)^{n}/n!~ + ~f(x)^k/k!}{\sum x^{n}/n!~ +~ x^k/k!} = 1$$

It implies P(k+1): $$~\lim_{x \to \infty} \frac{\sum_0^{k}f(x)^{n}/n!~ + ~f(x)^{k+1}/(k+1)!}{\sum x^{n}/n!~ +~ x^{k+1}/(k+1)!} = 1$$

Since P(k) implies P(k+1), and P(1) is true...?

Thanks.

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You can not use induction to prove a statement for all real numbers – Belgi Oct 2 '12 at 2:58
@Belgi That is not the source of the problem, his statement is about the functions f(x) and x, not about particular real numbers. The source of the error is that he has made a false conclusion after the induction - the induction itself is fine. Indeed, For any finite integer, the ratio of those polynomials is still $1.$ The Taylor series of $\exp$ is the limit of those polynomials and the induction says nothing about that case. Induction only gives you the result for every finite truncation of the taylor series. – Ragib Zaman Oct 2 '12 at 3:02
Things true for any prefix need not be true in the limit. For example, every prefix of the harmonic series is "convergent". To check what goes wrong in your example, you can try to work out what happens when $f(x)=x+1$. – Yuval Filmus Oct 2 '12 at 3:25
@daniel: A correct proof using induction does only prove the proposition for each finite case. If I tell you that I have functions $f_k(x)$ for $k=1,2,3,\ldots$, and that each $f_k(x)$ is continuous and bounded, you can't conclude anything about the pointwise limit "$f_{\infty}$", defined as $\lim_{k\rightarrow\infty}f_k(x)$, even assuming it exists. – mjqxxxx Oct 2 '12 at 3:39
@mjqxxxx: Ah. Yes this is I think easier for me to grasp. Thank you, that makes perfect sense (and is consistent with the above comments). Thank you. – daniel Oct 2 '12 at 3:43

This question has already been dealt with in the comments, but the following remark might also be helpful, since it relates it to the general phenomenon that a non-uniform limit of continuous functions need not be continuous:

Replacing $x$ by $1/x$ converts the limit under consideration into a limit as $x \to 0$ rather than $x \to \infty$. The question can then be reformulated more generally as follows: suppose that $f_n(x)$ is a sequence of functions on $(0,1]$ (say) such that $f_n(x) \to 1$ as $x \to 0$ for each value of $n$, and such that $\lim_{n \to \infty} f_n(x)$ exists for each $x \in (0,1]$; writing the limiting function as $f(x)$, is it then necessarily the case that $f(x) \to 1$ as $x \to 0$?

The answer is no, as is well-known. E.g. if $f_n(x) = 1 - x^{1/n}$, then $f(x) = \lim_{n \to \infty} 1 -x^{1/n} = 0$ for $x \in (0,1]$.

This is closely related to the fact that if we think of each $f_n(x)$ as a function on the closed interval $[0,1]$, then the $f_n(x)$ converge to $f(x)$ pointwise, but not uniformly, and the limiting function $f(x)$ (which takes the value $1$ at $x = 0$ but the value $0$ for all other $x$) is not continuous.

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This is a good amplification of mjqxxxx's note and again probably best addresses my confusion. Of course Ragib Zaman's answer is for sure the right answer. – daniel Oct 2 '12 at 12:39