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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 numbers2012-10-02
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    But the induction is on k, an integer?2012-10-02
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    @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.2012-10-02
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    I'm sure you are right (+1) but it's a subtle issue, to me. For any n you pick, however high, I pick an N > n for which this is true. In particular, we have 1 in S, k in S implies (k+1) in S. Then every positive integer is in S. In this case the set S is the set {0,1,2,...} of indices of in Taylor series...the phrase "every finite truncation" is difficult because "all truncations" are not much different from infinite truncations.2012-10-02
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    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 $f(x)=x+1$.2012-10-02
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    @YuvalFilmus: Right. That's how I proved to myself that the conclusion was false. But one could take a proof using induction that was correct, and assert that it only proved the proposition for each finite case.I think it's the English I find confusing.2012-10-02
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    Anyway thanks for the comments. I accept the answers and and will them over.2012-10-02
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    @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.2012-10-02
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    @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.2012-10-02
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    @daniel : Another way to see what goes wrong is that you're arguing somehow that $$ \lim_{n \to \infty} \lim_{x \to \infty} (thing / thing) = \lim_{x \to \infty} \lim_{n \to \infty} (thing / thing) = \lim_{x \to \infty} (e^{thing}/e^{thing}) $$ and it is not clear you can swap limits around. Because what you've proved is essentially that the limit in $x$ on the $LHS$ is always $1$, so you assumed you could "swap limits and conclude from the finite case that taking the limit in $n$ is okay".2012-10-02
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    Right, that is what I assumed, more or less. I saw right away that it was spurious but the induction troubled me--I think mjqxxxx has touched on the essential confusion. This is also helpful.2012-10-02

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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.2012-10-02