$\frac{1}{A}e^{(c-\varepsilon)\sqrt{n}} < g(n) < e^{c\sqrt{n}}$ implies $\log{g(n)} \sim c\sqrt{n}$.

Question

The following is from a research paper: Suppose $g(n) < e^{c\sqrt{n}}$ for some constant $c$ and for all integers $n \ge 1$. Suppose, in addition that for every $\varepsilon > 0$ there exists a constant $A > 0$ such that $g(n) > \frac{1}{A}e^{(c-\varepsilon)\sqrt{n}}$. They concluded that $\log{g(n)} \sim c\sqrt{n}$. I can see that for a fixed $\varepsilon > 0$,

$$ \begin{align*} \frac{1}{A}e^{(c-\varepsilon)\sqrt{n}} &< g(n) < e^{c\sqrt{n}} \\ (c-\varepsilon)\sqrt{n} - \log A &< \log g(n) < c\sqrt{n} \\ \frac{(c-\varepsilon)}{c} - \frac{\log A}{c\sqrt{n}} &< \frac{\log g(n)}{c\sqrt{n}} < 1 \end{align*} $$ As $n \rightarrow \infty$, the left term in the inequality goes to $\frac{c-\varepsilon}{c}$. I know this can be made arbitrarily close to 1 by taking $\varepsilon$ arbitrarily small, but because of the dependence of the constant $A$ on epsilon, I feel like I can't conclude anything. Some clarification will be appreciated.

Answer 1

So, let's extract the core question here:

Suppose that there exist two functions $f,\delta\colon \mathbb{N}\to \mathbb{R}$ such that $\lim_{n\to\infty} \delta(n)=0$ and, for every $\varepsilon >0$, there exists $A_\varepsilon$ such that $$ 1-\varepsilon - A_\varepsilon\cdot \delta(n) \leq f(n) \leq 1 \tag{1} $$ Then do we have $\lim_{n\to\infty} f(n)=1$?

Note that your question correcponds to $f(n) = \frac{\log g(n)}{c\sqrt{n}}$; and the conclusion is exactly the definition of $\operatorname*{\sim}_{n\to\infty}$. Fix any $\varepsilon>0$. From (1), we get (taking $n\to\infty$) that $$ \lim\!\inf_{n\to\infty} f(n) \geq 1-\varepsilon \tag{2} $$ Since this holds for every $\varepsilon>0$, we can take the limit as $\varepsilon \searrow 0$ to get $$ \lim\!\inf_{n\to\infty} f(n) \geq 1 \tag{3} $$ and since $f(n) \leq 1$ by (1), we get $\lim_{n\to\infty} f(n) = 1$.