Let $\displaystyle f(x)= e^{-Ae^x} e^{Bx} \left( 1+ \mathcal{O} \left( e^{-x} \right)\right)$, where $A$ and $B$ are constants.
Claim: $|f(x)| = \mathcal{O}\left( \exp(-e^x)\right)$ as $x \rightarrow \infty$.
Note that $$\frac{1}{e^{Ae^x}} \leq \frac{e^{Bx}}{e^{Ae^x}} \leq \frac{e^{x}}{e^{Ax}}\leq \frac{1}{(e^x)^{A-1}}$$ hence $$\lim_{x\rightarrow \infty} \frac{e^{Bx}}{e^{Ae^x}}=0,$$ so $e^{Ae^{x}}$ grows much faster than $e^{Bx}$. Is my argument here correct? I feel like i've done something wrong with my argument or missing something or is my claim even holds? Any comment and help would be so much appreciated! Thanks in advance!