The value of $$\lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{2n-1} \frac{n^2}{n^2 + 4r^2}$$ is:

We have to evaluate the limit

$$\lim_{n \to \infty}\;\frac{1}{n}\sum_{r=0}^{2n-1}\frac{n^{2}}{n^{2}+4r^{2}}.$$

First, write the summand in a form that exposes a factor of $$n^{2}$$ in the denominator:

$$\frac{n^{2}}{n^{2}+4r^{2}} =\frac{n^{2}}{n^{2}\Bigl(1+\dfrac{4r^{2}}{n^{2}}\Bigr)} =\frac{1}{1+4\dfrac{r^{2}}{n^{2}}}.$$

Now introduce the new variable

$$x=\frac{r}{n}\qquad\Longrightarrow\qquad r=nx,$$

so that as $$r$$ increases by $$1$$, the corresponding $$x$$ increases by

$$\Delta x=\frac{1}{n}.$$

With this substitution, the fraction becomes

$$\frac{1}{1+4x^{2}},$$

because $$\dfrac{r^{2}}{n^{2}}=x^{2}.$$

Hence the entire expression takes the Riemann-sum form

$$\frac{1}{n}\sum_{r=0}^{2n-1}\frac{n^{2}}{n^{2}+4r^{2}} =\sum_{r=0}^{2n-1}\frac{1}{1+4x_{r}^{2}}\;\Delta x,$$

where $$x_{r}=\dfrac{r}{n}$$ and $$\Delta x=\dfrac{1}{n}.$$

As $$n\to\infty,$$ the points $$x_{r}$$ run through the entire interval from $$0$$ up to $$2$$ (because $$r$$ goes from $$0$$ to $$2n-1$$), and the width $$\Delta x$$ shrinks to zero. Therefore the limit of the Riemann sum equals the definite integral

$$\int_{0}^{2}\frac{1}{1+4x^{2}}\;dx.$$

Now we evaluate this integral. We recall the standard formula

$$\int\frac{dx}{a^{2}+x^{2}}=\frac{1}{a}\tan^{-1}\!\Bigl(\frac{x}{a}\Bigr)+C,$$

and notice that $$1+4x^{2}=1+(2x)^{2}.$$ Hence we put $$t=2x\;( \text{so }dt=2\,dx \;\text{and}\;dx=\dfrac{dt}{2} ),$$ obtaining

$$\int\frac{1}{1+(2x)^{2}}\;dx =\int\frac{1}{1+t^{2}}\;\frac{dt}{2} =\frac{1}{2}\tan^{-1}(t)+C =\frac{1}{2}\tan^{-1}(2x)+C.$$

Applying the limits $$x=0$$ to $$x=2$$ we get

$$\int_{0}^{2}\frac{dx}{1+4x^{2}} =\left[\frac{1}{2}\tan^{-1}(2x)\right]_{0}^{2} =\frac{1}{2}\Bigl(\tan^{-1}(4)-\tan^{-1}(0)\Bigr) =\frac{1}{2}\tan^{-1}(4),$$

because $$\tan^{-1}(0)=0.$$

Therefore

$$\lim_{n \to \infty}\;\frac{1}{n}\sum_{r=0}^{2n-1}\frac{n^{2}}{n^{2}+4r^{2}} =\frac{1}{2}\tan^{-1}(4).$$

Comparing with the given options, this matches Option C.

Hence, the correct answer is Option C.