Solution
= Solution
{c}
The ratio can be evaluated directly:
$$
\frac{e^{f(x)}}{e^{g(x)}}=\exp(f(x)-g(x)).
$$
The assumption $|f-g|\to0$ gives $f-g\to0$, so <continuity> of the <exponential function> gives
$$
\boxed{e^{f(x)}\sim e^{g(x)}.}
$$
The additional premise $f\sim g$ is not needed for this implication. The useful requirement is additive control of the exponents, as explained by <asymptotic comparison of exponential functions>.