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.

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 $M(n) \to \frac{3+\sqrt{5}}{2}$.
