Consider two rational numbers $a, b > 0$. What can be said of the following limit as a function of $a, b$, where $x$ and $y$ are natural numbers?
$$ L = \lim_{(x,y)\to\infty}\dfrac{ \displaystyle\binom{x + y}{x} }{ \displaystyle\binom{ax + by}{ax} } $$
For reasons that I will show below, I make two claims:
Claim 1.
- If both $a, b > 1$ then $L = 0$
- If both $a, b < 1$ then $L = \infty$
Claim 2. If $a < 1$ and $b > 1$ (or vice versa), the limit is undefined, i.e., there are two different sequences of $(x,y)$ approaching infinity such that the first yields $0$ and the second yields $\infty$.
The question: Are the claims above true? (If yes, how to prove them?)
Note: Since both $x$ and $y$ are going to infinity, we can assume that both $ax$ and $by$ are integers. Also, the edge case of $a = b = 1$ can be ignored.
My reasoning
Define $f(u, v) = \displaystyle\binom{u + v}{u} = \displaystyle\binom{u + v}{v}$. Applying the approximation $\displaystyle\binom{n}{k} \approx \dfrac{n^k}{k!}$, we get:
$$f(u, v) \approx \dfrac{(u+v)^u}{u!} \approx \dfrac{(u+v)^v}{v!}$$
Now, assuming that only $x$ goes to infinity:
$$\begin{align} L_X &= \lim_{x\to\infty} \dfrac{f(x,y)}{f(ax,by)} \\\\&= \lim_{x\to\infty} \dfrac{ \dfrac{(x+y)^y}{y!} }{ \dfrac{(ax+by)^{by}}{(by)!} } \\\\&= \lim_{x\to\infty} \dfrac{ (x+y)^y }{ (ax+by)^{by} } \\\\&= \lim_{x\to\infty} \dfrac{x^y}{(ax)^{by}} \\\\&= \lim_{x\to\infty} \dfrac{x^y}{a^{by}x^{by}} \\\\&= \lim_{x\to\infty} \dfrac{x^y}{x^{by}} \\\\&= \lim_{x\to\infty} x^{y(1-b)} \\\\&= \left\{ \begin{array}{c} \infty \quad \text{ if $b$ < 1} \\ 0\,\,\, \quad \text{ if $b$ > 1} \end{array} \right. \end{align}$$
Analogously, assuming that only $y$ goes to infinity, we get:
$$ L_Y = \lim_{y\to\infty} \dfrac{f(x,y)}{f(ax,by)} = \left\{ \begin{array}{c} \infty \quad \text{ if $a$ < 1} \\ 0\,\,\, \quad \text{ if $a$ > 1} \end{array} \right. $$
The observations above suggest to me, intuitively, that $a > 1$ "pulls" $L$ to zero and so do $b > 1$. This way, if both $a, b > 1$, probably $L = 0$ and if both $a, b < 1$, probably $L = \infty$.
And, in turn, if one of them is greater than $1$ and the other is less than $1$, probably it is possible to choose a certain way to approach infinity that will yield $0$, and another way that will yield $\infty$, therefore making $L$ itself undefined.