, 2018

Let us first analyze $x_n$ as $n \to \infty$.

We use the Taylor expansion for $e^x$ near $x = 0$:
$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$$
So,
$$e^{1/n} - 1 = \frac{1}{n} + \frac{1}{2n^2} + \frac{1}{6n^3} + O\left(\frac{1}{n^4}\right)$$
$$e^{1/n^2} - 1 = \frac{1}{n^2} + \frac{1}{2n^4} + \frac{1}{6n^6} + O\left(\frac{1}{n^8}\right)$$
Therefore,
$$\frac{e^{1/n} - 1}{e^{1/n^2} - 1} = \frac{\frac{1}{n} + \frac{1}{2n^2} + \frac{1}{6n^3} + O\left(\frac{1}{n^4}\right)}{\frac{1}{n^2} + \frac{1}{2n^4} + O\left(\frac{1}{n^6}\right)}$$
Divide numerator and denominator by $1/n^2$:
$$= \frac{n + \frac{1}{2} + \frac{1}{6n} + O\left(\frac{1}{n^2}\right)}{1 + \frac{1}{2n^2} + O\left(\frac{1}{n^4}\right)}$$
Now, expand $1/(1 + \varepsilon)$ for small $\varepsilon$:
$$\frac{n + \frac{1}{2} + \frac{1}{6n} + O\left(\frac{1}{n^2}\right)}{1 + \frac{1}{2n^2} + O\left(\frac{1}{n^4}\right)} = \left(n + \frac{1}{2} + \frac{1}{6n}\right)\left(1 - \frac{1}{2n^2} + O\left(\frac{1}{n^4}\right)\right)$$
Multiply out:
$$= n + \frac{1}{2} + \frac{1}{6n} - \frac{n}{2n^2} - \frac{1}{4n^2} - \frac{1}{12n^3} + O\left(\frac{1}{n^2}\right)$$
$$= n + \frac{1}{2} + \frac{1}{6n} - \frac{1}{2n} - \frac{1}{4n^2} - \frac{1}{12n^3} + O\left(\frac{1}{n^2}\right)$$
Now, $x_n = \frac{e^{1/n} - 1}{e^{1/n^2} - 1} - n$:
$$x_n = \frac{1}{2} + \frac{1}{6n} - \frac{1}{2n} - \frac{1}{4n^2} - \frac{1}{12n^3} + O\left(\frac{1}{n^2}\right)$$
$$x_n = \frac{1}{2} - \frac{1}{3n} - \frac{1}{4n^2} + O\left(\frac{1}{n^2}\right)$$
Therefore,
$$\lim_{n \to \infty} x_n = \frac{1}{2}$$

Now, compute $\lim_{n \to \infty} n(x_n - 1/2)$:
$$n(x_n - 1/2) = n\left(-\frac{1}{3n} - \frac{1}{4n^2} + O\left(\frac{1}{n^2}\right)\right) = -\frac{1}{3} - \frac{1}{4n} + O\left(\frac{1}{n}\right)$$
So,
$$\lim_{n \to \infty} n(x_n - 1/2) = -\frac{1}{3}$$