interpreting limits

Question

a short question:

is this true that if:

f(x) = $2^x$
g(x) = $100^\sqrt{x}$

$\lim\limits_{x \to \infty} \dfrac{f(x)}{g(x)}$ = $\infty$

then for x sufficiently large f(x) is always greater than g(x)?

And if it is true, is it sufficient to claim that

g(x) $\in$ O(f(x)) ?

Answer 1

Note that $$2^x=2^{\frac{x}{2}+\frac{x}{2}}=(2^{\frac{\sqrt{x}}{2}})^{\sqrt{x}}2^{\frac{x}{2}}.$$

If $x$ is large enough, then $2^{\frac{\sqrt{x}}{2}}\gt 100$, and therefore $(2^{\frac{\sqrt{x}}{2}})^{\sqrt{x}}\gt 100^{\sqrt{x}}$.

It follows that for large enough $x$, we have $\frac{100^{\sqrt{x}}}{2^x}\gt 2^{\frac{x}{2}}$, and therefore the ratio blows up (fast).

We can conclude that $g(x)\in O(f(x))$, and indeed that $g(x)\in o(f(x))$.

Remark: For no particularly good reason, I prefer to consider the ratio $\frac{g(x)}{f(x)}$, and show that it is bounded above. In this case, we have $\lim_{x\to\infty}\frac{g(x)}{f(x)}=0$, so the ratio is certainly bounded above.

Answer 2

True because $2^x = (2^{x^{1/2}})^{x^{1/2}} > 100^{x^{1/2}}$ when x is big enough.

g(x) = o(f(x)) and not O(f(x)).