Dtyjlui

as following a proof from a book it state as(please see pic if still confusing):

since lim ->∞ (f(n)/g(n)) = c >0, it follows from the definition of a limit that there is some n0 beyond which the ration is always between 1/2 c and 2c. Thus, f(n) <2cg(n) for all n >= n0, which implies that f(n) = O(g(n));

Can someone please explain a bit why there is some n0 beyond which the ratio is always between 1/2c and 2c ? enter image description here

That's the very definition of $lim_{nto infty} h(n) = c$.

That means for any value $epsilon > 0$ we can find some value $n_0$ so that for all $n > n_0$ ir will always be then case that $c-epsilon < h(n) < c +epsilon$.

There's nothing to explain why, because that is what $lim_{nto infty} h(n) = c$ means.

So $lim_{nto infty} frac {f(n)}{g(n)} = c$ then we can find for ANY $epsilon > 0$, in particular for $epsilon = frac 12 c$, there is some $n_0$ so that for all $n > n_0$ we will always have:

$c - epsilon < frac {f(n)}{g(n)} < c + epsilon$ so

$c - frac 12 c < frac {f(n)}{g(n)} < c + frac 12c $ so

$frac 12 c < frac {f(n)}{g(n)} < frac 32c < 2c$

This is just a quick and dirty version of an $epsilontext{-}delta$ argument; the fact that $c>0$ means that $epsilon_1=c/2>0$ and $epsilon_2=c>0$ are both valid "choices" for the limit definition, and then the rest falls into place.