Calculate this awesome limit!

Let the radius of the circle be $r$ and the side length of the equivalent triangle containing $n$ rows of circles be $a$ . From the figure, we note that $a=2r(n-1)+2\sqrt 3r$ $= 2r(n-1+\sqrt 3)$ . Then the area of the triangle is $A = \dfrac 12 a \times \dfrac {\sqrt 3}2a$ $= \sqrt 3r^2(n-1+\sqrt 3)^2$ .

For $n$ rows of circles, there are $\displaystyle \sum_{k=1}^n k = \dfrac {n(n+1)}2$ circles. Therefore, the area of all the circles $B = \dfrac {n(n+1)\pi r^2}2$ .

Now we have:

$\begin{aligned} \lim_{n \to \infty} \frac BA & = \lim_{n \to \infty} \frac {n(n+1)\pi \cancel{r^2}}{2\sqrt 3 \cancel{r^2}(n-1+\sqrt 3)^2} & \small \color{#3D99F6} \text{Divide up and down by }n^2 \\ & = \lim_{n \to \infty} \frac {\left(1+\frac 1n\right)\pi}{2\sqrt 3\left(1-\frac {1-\sqrt 3}n\right)^2} \\ & = \frac {\bigg(1+ \cancel{\frac 1n}^0\bigg)\pi}{2\sqrt 3\bigg(1-\cancel{\frac {1-\sqrt 3}n}^0\bigg)^2} \\ & = \frac \pi {2\sqrt 3} \approx \boxed{0.907} \end{aligned}$

We can ignore the circles that are tangent to the triangle(the outer layer of circles),then it is as same as a circle inscribed in a regular hexagon,so it's $\dfrac{\pi}{2\sqrt 3}$ (We could cut the triangle in this way,and look at the circle in the middle.It's inscribed in a hexagon.)

(Though we should ignore 9 circles when $n=4$ ,but when $n$ gets bigger,we can actually ignore it because the circles we ignore $=3n-3$ ,but the total of the triangle $=\dfrac{n(n+1)}2$ )

Let us call each circle has radius as $R$ . The total ammount of circles $T$ will be:

$\large \displaystyle T = \sum_{i=1}^n i =\frac{n^2+n}{2}$

So:

$\large \displaystyle B = T \cdot \pi R^2$

$\color{#20A900} \boxed{\large \displaystyle B = \frac{\pi R^2 (n^2 + n)}{2} }$

Let us call the triangle side as $S$ . Looking at the base, the distance from the center of the leftmost circle to the center of the rightmost circle is $2R \cdot (n - 1)$ . The part at the left of center of the leftmost circle or at the right ot the center of the rightmost circle is one cathetus of a right triangle, adjacent to a side which angle is $30^o$ and with other cathetus equal to $R$ . So, this part is equal to $R \sqrt{3}$ and, since there are two of these parts:

$\large \displaystyle S = 2R(n-1 + \sqrt{3} )$

So:

$\large \displaystyle A = \frac{S^2 \sqrt{3}}{4}$

$\color{#20A900} \boxed{ \large \displaystyle A = R^2 \left [ n^2 \sqrt{3} + (6 - 2 \sqrt{3}) n + (4 \sqrt{3} - 6) \right ] }$

Thus:

$\large \displaystyle L = \lim_{n \rightarrow \infty} \frac{B}{A}$

$\large \displaystyle L = \lim_{n \rightarrow \infty} \frac{\pi R^2 (n^2 + n)}{2R^2 \left [ n^2 \sqrt{3} + (6 - 2 \sqrt{3}) n + (4 \sqrt{3} - 6) \right ]}$

$\color{#3D99F6} \boxed{ \large \displaystyle L = \frac{\pi}{2 \sqrt{3}} \approx 0.907}$

In limiting case n (2r) = l. So, r = l/(2n) . Area of all circles/Total area =(appx) (pi /8)/(root(3)/4) = pi/(2 root(3))

Chew-Seong Cheong's solution is close the expected solution I started off with!