Find whether $f(x)$ is $O(g(x))$ [whether $f(x)$ is Big-O of $g(x)$]

Question

Given: $f(x)=3^{\sqrt{x}}, g(x) = 2^x$, find whether $f(x)$ is Big-O of $g(x)$, and vice-versa.

I want to use the following fact: $$\lim_{x\to\infty}(\ln|f(x)|-\ln|g(x)|) \leq ln(C) \implies f(x)=O[g(x)]$$

I have done the following: $$\lim_{x\to\infty}(\ln|f(x)|-\ln|g(x)|)$$ $$ = \lim_{x\to\infty}(\sqrt{x}\ln(3)-xln(2))$$ $$ = \lim_{x\to\infty}[\ln(\dfrac{3^\sqrt{x}}{2^x})] $$ $$= \ln[\lim_{x\to\infty}(\dfrac{3^\sqrt{x}}{2^x})] $$

Where do I go from here?

Answer 1

You want to compare $f(x)$ and $g(x)$

Let's take the logarithm of their ratio

$\ln\left(\dfrac{f(x)}{g(x)}\right)=\ln(f(x))-\ln(g(x))=\sqrt{x}\ln(3)-x\ln(2)\sim-x\ln(2)\to-\infty$

Since the square root of $x$ is negligible compared to $x$ at infinity: $\sqrt{x}\ll x$

Thus $\dfrac{f(x)}{g(x)}\to 0$ and $f(x)=o(g(x))$ which is even stronger than just a big-O.

Anyway it implies $f(x)=O(g(x))$

Answer 2

We have that

$$\large 3^{\sqrt{x}}=2^{\sqrt x \log_2 3} \implies \frac{3^{\sqrt{x}}}{2^x}=2^{\sqrt x \log_2 3-x}\to 0$$

Answer 3

You can write $f(x)=3^{\sqrt x}=2^{\sqrt xlog_23}$.

Now $ln(\lim\limits_{x\to \infty}(3^\sqrt x/2^x) )=log_2(\lim\limits_{x\to \infty}(2^{\sqrt xlog_23}/2^x) )\leq log_2(C)$

So ${\lim\limits_{x\to \infty}\sqrt xlog_23-x\leq C}$

$\sqrt xlog_23\leq x+C$ as $x \to \infty$

$2^{\sqrt xlog_23}\leq 2^{x+C}$ as $x\to \infty$

$\implies f(x)\leq 2^Cg(x)$

Hence $f(x)=O(g(x))$ by definition.