 Solution

Take $g(x)=x$ and $f(x)=x+1$. Then $f/g=1+1/x\to 1$, so $f\sim g$. However,for every $x$, and this ratio does not approach one. Thus asymptotic equivalence need not survive exponentiation. A small relative error between large exponents can leave a nonvanishing additive error. This is the distinction in asymptotic comparison of exponential functions.