Prove or disprove: if $f(n)= o(g(n))$ and $h=\omega(1)$ then, $f(h(n))= o(g(h(n)))$
ive tried to approach it with limits calculation, but no matter which arguments I used the equation seems to be proven.
ex:
Let $f(x) = n$, $g(x) = n^2$, and $h(x)=2^n$.
$\lim_{x\to \infty} = \cfrac{f(x)}{g(x)}= \cfrac{n}{n^2}=0$
$\lim_{x\to \infty} = \cfrac{h(x)}{1}= 2^n=\infty$
$\lim_{x\to \infty} = \cfrac{f(h(x))}{g(h(x))}= \cfrac{2^n}{2^{n^2}}=2^{n-n^2}=2^{n^2(\cfrac{1}{n}-1)}=2^{-\infty} = 0$
how do I prove or disprove this kind of question properly?
Well if you are free to assume that they are general functions, then you can take:
$$ f(x)=\begin{cases} x & ;x\in \mathbb{N} \\ 1 & ;\text{else} \end{cases} \qquad g(x)= \begin{cases} x^2 & ;x\in \mathbb{N} \\ 1 & ;\text{else} \end{cases} \qquad \text{and} \qquad h(x)=x+\frac{1}{2} $$
Then $h(n)=n+\frac{1}{2}$ and:
$$ \frac{f\big(h(n) \big)}{g\big( h(n) \big)} \equiv 1 $$
The condition only gives you an idea how $f$ and $g$ behave relative to one another where the argument is a natural number. But you seem to have freedom with regards to how they behave elsewhere. That's my idea at least.
I can add though that if $h(n)\in \mathbb{N}$ for all $n$, then the relation must hold since $\frac{f\circ h(n)}{g\circ h(n)}$ is a subseqeunce of $\frac{f(n)}{h(n)}$ which must converge to the same limit.