Show that $\lim_{n\to\infty}f(n)/g(n)= 0 \implies \lim_{n\to\infty}3^{f(n)}/3^{g(n)} = 0?$ when both functions are monotone growing and positive

Question

$f(n), g(n)$ are both monotone growing functions such that $f(n), g(n) > 1 \ \forall n $.

How can I prove that $$\lim_{n\to\infty}\frac{f(n)}{g(n)}= 0 \implies \lim_{n\to\infty}\frac{3^{f(n)}}{3^{g(n)}} = 0?$$

Answer 1

Detailed outline: write $f(n) = \varepsilon(n)g(n)$, with by assumption $\lim_{n\to\infty}\varepsilon(n) = 0$. Then $$ \frac{3^{f(n)}}{3^{g(n)}} = \frac{1}{3^{g(n)-f(n)}} = \frac{1}{3^{(1-\varepsilon(n))g(n)}}\,.\tag{1} $$ Since $x\mapsto 3^x$ is continuous, it suffices to prove that $$ \lim_{n\to\infty}(1-\varepsilon(n))g(n) = +\infty\,.\tag{2} $$ Since $\lim_{n\to\infty}(1-\varepsilon(n)) = 1$, it suffices to prove that $$ \lim_{n\to\infty} g(n) = +\infty\,.\tag{3} $$ But this follows from our assumption that $f(n)\geq 1$ for all $n$: $$ g(n) \geq \frac{g(n)}{f(n)} = \frac{1}{f(n)/g(n)} \xrightarrow[n\to\infty]{} \infty\,. $$

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Note that this would be false without the monotonicity assumption on $f,g$, as shown in this other question.