Given a random $X$ of fair binomial distribution for $n$ large, what are asymptotes for $P(X\geq k)$ and $P(X=k)/P(X\geq k)?$ Given a random variable $X\sim \operatorname{Binom}(n,1/2).$ With $n$ large and $\sqrt{n}<k<n-\sqrt{n},$ what are asymptotes for the probability $$P(X\geq k)$$ and the conditional probability $$P(X=k\mid X\geq k)=\frac{P(X=k)}{P(X\geq k)}?$$
Feel free to pick a tighter range for $k,$ if needed for the asymptote.
If this is hard, perhaps, given $\mu\in[0,1]$ and large $n,$ for the smallest $k$ such $$P(X=k\mid X\geq k)\geq \mu$$ estimate $P(X\geq k).$
In this latter formulation, we can restrict to $|\mu-1/2|\leq\frac1{\sqrt{n}}.$
I know when $n=2k-1,$ $$P(X\geq k)=\frac1{2}\\P(X=k\mid X\geq k)\sim\frac{2}{\sqrt{k\pi}}$$ and when $n=2k,$ $$P(X\geq k)-\frac12\sim \frac{1}{2\sqrt{k\pi }}\\P(X=k\mid X\geq k)\sim \frac{2}{\sqrt{k\pi}+1}
$$
This showed up while working on this (very partial) answer.
I’ve got enough probability in my head to know that it has something to do with the way that the binomial distributions converge to the normal distribution, but other than that, I got nothing.
I didn’t find anything obviously useful on Wikipedia for the binomial distribution, but I may have missed something.
 A: Without continuity corrections, using the CLT, so that $X \to \mathcal{N}(\mu, \sigma^2)$ with $\mu=n/2$ and $\sigma^2 = n/4$ we have
$$P(X=k)\approx \sqrt{\frac{2}{\pi n}} \exp\left( -   \frac{(n-2k)^2}{2n }\right)$$
$$P(X \ge k)\approx 1- \Phi\left( \frac{2k -n}{\sqrt n}\right)$$
where  $\Phi$ denotes the standard Gaussian CDF.
Your range $\sqrt{n}<k<n-\sqrt{n}$ is quite large, I'm afraid there's not much more to symplify here.

Edit: The above approximation looks quite good for both the numerator and denominator, but is practically useless for computing the ratio. More care is needed. At least the continuity correction is required for the denominator. Then we get
$$r_{n,k} = \frac{P(X=k)}{P(X\geq k)} \approx \frac{\phi(z)}{1-\Phi(z-\frac{1}{2\sigma})} \tag 1$$
where $z= (k-\mu)/\sigma$, $\mu = n/2$, $\sigma = \sqrt{n}/2$.
This first order approximation is acceptable not very far from the mean (say, $\frac12 n \pm \frac32 \sqrt{n}$)
Trying to approximate $(1)$ by the Mill's ratio, as suggested in the comments, does not seems promising, because that approximation is for the tails, and $(1)$ is already bad for the tails.
Edit 2:
For the upper tail ($ k \to n$) and small $j\ge 0$ I get
$$ \frac{\binom{n}{k+j}}{\binom{n}{k}} \approx \left(\frac{n}{k}-1\right)^j \tag 2$$
and then
$$r_{n,k} \approx  2 - \frac{n}{k} \tag 3$$
The graph shows, for $n=70$ the exact value in blue, and the first order approximation $(1)$ in yellow. Tail approximation $(3)$ in green. The shaded region corresponds to the $3-$sigma interval.

