"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. 
 A: 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.
