Is it true that $$ \lim_{n\to\infty} \frac{\Omega(n)}{\omega(n)} = 1$$ where $\Omega$ counts the number of prime factors of $n$ and $\omega$ counts the number of distinct prime factors of $n$.
Is it true that $$\omega(n) \sim \Omega(n)$$
Where $$\omega(n)$$and$$\Omega(n)$$ are the prime omega functions
No, it's not true. As Robert Israel's question comment suggests, consider $n = 2^m$ for $m \ge 1$. You then get
$$\frac{\Omega(n)}{\omega(n)} = \frac{m}{1} = m \tag{1}\label{eq1A}$$
As such, since $m$ can go to infinity, there's no upper bound on the fraction, so its limit can't be $1$. In fact, for primes, for example, the fraction is $1$, plus you can also get other larger values also depending on the value of $n$, so there's actually no limit of that fraction.