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.