Limit of summation $\left(-1\right)^{x}\sum_{r=0}^{x}\left(-1\right)^{r}\binom{x}{r}\left(\frac{r}{x}\right)^{t(x)}$ I am trying to find what the function $t(x)$ would have to be in order to make the expression true.
$$\lim_{x \to \infty} \left(-1\right)^{x}\sum_{r=0}^{x}\left(-1\right)^{r}\binom{x}{r}\left(\frac{r}{x}\right)^{t(x)}=\frac{1}{2}$$
I've tried using Stirling's approximation to see if things cancelled out to get something simpler to solve but it ended up looking worse. From there I haven't know where to go next.
 A: For $t(x) = x(\ln(x)+\alpha)$, the limit is $e^{-e^{-\alpha}}$. For $\alpha = e^{-1}$, the limit is $\approx .50047$, explaining Lewis's finding. For $\alpha = -\ln(\ln(2))$, the limit is $1/2$. So $t(x) = x\ln(x/\ln(2))$ works.
For the proof, change $r$ to $x-r$ to get $\sum_{r=0}^x (-1)^r {x \choose r}(1-\frac{r}{x})^{t(x)}$. We can clearly rid of the terms $r > 10x/\ln(x)$ (for large enough $x$) since ${x \choose r} \le 2^x$ and $(1-\frac{r}{x})^{t(x)} \le \exp(-5x)$. When $r \le 10x/\ln(x)$, we have $(1-\frac{r}{x})^{t(x)} = \exp\left(-t(x)\ln(\frac{1}{1-\frac{r}{x}})\right) = \exp\left(-r\frac{t(x)}{x}+O(t(x)\frac{r^2}{x^2})\right) = (1+o(1))\exp\left(-r(\ln(x)+\alpha)\right)$. Therefore, our sum has the same limit as $\sum_{r=0}^{10x/\ln(x)} (-1)^r {x \choose r} \exp(-r(\ln(x)+\alpha))$. But since $r > 10x/\ln(x)$ has $\exp(-r(\ln(x)+\alpha)) \le \exp(-5x)$, we can replace by the complete sum $\sum_{r=0}^x (-1)^r {x \choose r} \exp\left(-r(\ln(x)+\alpha)\right)$. By the binomial theorem, this sum is $(1-e^{-(\ln(x)+\alpha)})^x = (1-\frac{e^{-\alpha}}{x})^x$, which converges to $e^{-e^{-\alpha}}$.
