$$\displaystyle\lim_{n \to \infty} \dfrac{1}{2n}\left(\dfrac{1}{\sqrt{1 - \frac{1}{2n}}} + \dfrac{1}{\sqrt{1 - \frac{2}{2n}}} + \dfrac{1}{\sqrt{1 - \frac{3}{2n}}} + \ldots + \dfrac{1}{\sqrt{1 - \frac{2n-1}{2n}}}\right)$$ is equal to

We need to evaluate:

$$ \lim_{n \to \infty} \frac{1}{2n}\left(\frac{1}{\sqrt{1 - \frac{1}{2n}}} + \frac{1}{\sqrt{1 - \frac{2}{2n}}} + \cdots + \frac{1}{\sqrt{1 - \frac{2n-1}{2n}}}\right) $$

Setting $$N = 2n$$ transforms the expression into

$$ \frac{1}{N}\sum_{k=1}^{N-1} \frac{1}{\sqrt{1 - \frac{k}{N}}} $$

This sum can be recognized as a Riemann sum for the function $$f(x) = \dfrac{1}{\sqrt{1-x}}$$ on the interval $$[0,1]$$, taking sample points $$x_k = \dfrac{k}{N}$$ with step size $$\Delta x = \dfrac{1}{N}$$. Although $$f(x)$$ becomes unbounded as $$x\to1$$, the corresponding improper integral converges, and thus the Riemann sum converges to the integral in the limit as $$N\to\infty$$:

$$ \lim_{N \to \infty} \frac{1}{N}\sum_{k=1}^{N-1} \frac{1}{\sqrt{1-k/N}} = \int_0^1 \frac{1}{\sqrt{1-x}}\,dx $$

To evaluate the integral, substitute $$u = 1-x$$ so that $$du = -dx$$. This gives

$$ \int_0^1 \frac{dx}{\sqrt{1-x}} = \int_1^0 \frac{-du}{\sqrt{u}} = \int_0^1 u^{-1/2}\,du = \left[2\sqrt{u}\right]_0^1 = 2(1)-2(0) = 2 $$

The answer is Option C: $$2$$.