Find $\lim_{n \to \infty} M(n)/n$ where $M(n)$ is the largest integer satisfying a binomial inequality Let $M (n) $ be the largest integer such that  ${m \choose n-1}>{m-1 \choose n} $, then what is the value of $\lim _{n \to \infty} \frac {M (n)}{n} $?
Attempt: After solving and simplifying the given condition, I get a quadratic in $m$ as $m^2-m (1+3n)+n^2-n < 0$. Solving for $m$ we have $m=\frac {(3n+1) \pm  \sqrt {5n^2+10n+1}}{2} $ thus I get two values for limit as $\frac{3 \pm \sqrt {5}}{2} $ . As both are positive I can't discard any value straight forward. So what should be done to discard one value or are both values the answer? Note that there can be  more than one correct  answer.
 A: Note that from the condition we get the ratio of the two binomials to be $\frac{mn}{(m-n+1)(m-n)}$ so our condition simplifies to $(m-n+1)(m-n) - mn <0$. That is $m^2 - 3mn + m - n + n^2< 0$. (note that this is different to your condition). We can simplify this to $m^2 - m(3n-1) + n^2 - n < 0$. This quadratic has roots $$\alpha,\beta = \frac{3n-1 \pm \sqrt{5n^2 - 2n+1}}{2}$$ where $\alpha$ corresponds to the root larger root and $\beta$ the smaller one. (Note that this differs from your roots as well). 
It is clear that $m$ must lie between $\beta < m < \alpha$. Since $\alpha - \beta= \sqrt{5n^2 - 2n +1} > 1$ then there is at least one integer between $\beta$ and $\alpha$. So $M(n)$ is the greatest integer in this interval. That is $\alpha - 1 \leq M(n) \leq \alpha$ or equivalently $\frac{\alpha - 1}{n} \leq \frac{M(n)}{n} \leq \frac{\alpha}{n}$.
Now we squeeze $$\lim \frac{\alpha - 1}{n} \leq \frac{\lim M(n)}{n} \leq \lim \frac{\alpha}{n}$$
But since $$\lim \frac{\alpha}{n} = \lim\frac{3 - \frac{1}{n} + \sqrt{5 - \frac{2}{n} + \frac{1}{n^2}}}{2} = \frac{3+\sqrt{5}}{2}$$ and $\lim \frac{\alpha - 1}{n} = \lim \frac{\alpha}{n}$ we squeeze $M(n) \to \frac{3+\sqrt{5}}{2}$.
