Is this conclusion on orders of magnitude correct?

Question

Let $f(n_1,n_2) = \mathcal{O}\left(\frac{n_1n_2^2}{(n_1-n_2)^3}\right)$, where $n_1$ and $n_2$ are natural numbers. If $n_1\propto n_2$, that is, if the two variables grow proportionally, is it true that $f(n_1,n_2) =\mathcal{O}(1)$?

Answer 1

It depends on the coefficient of proportionality. If it is different from $1$, yes. If not, $f$ will tend to infinity.

For instance, if $n_1(k) = k$, and $n_2(k) = k-1$, it is easy to see that $\frac{n_1n_2^2}{(n_1-n_2)^3} \simeq k^3$ as $k$ tends to infinity.

However, if, say $n_1(k) \geq (1+\delta) n_2(k)$, where $\delta >0$, this cannot happen.