Limit with Sphere

Let $r$ and $h$ be the radius of the base and the height of the spherical cap, respectively. Then from the right triangle shown in the diagram in the question, we get

$\begin{array}{rl} r^2 &= \ \ 1^2 - (1 - h)^2 \\ &= \ \ 2h - h^2 \end{array}$

The area of a spherical cap is given by $2 \pi R h$ , where $R$ is the radius of the sphere; in our case we have $R = 1$ , so $A(a) = 2 \pi h$ . Also, $a = \pi r^2 = \pi (2h - h^2)$ .

Then, since as $a \to 0$ we have $h \to 0$ , we get

$\begin{array}{rl} \lim_{a \to 0} \dfrac{A(a) - a}{a^2} &= \ \ \lim_{h \to 0} \dfrac{2 \pi h - \pi (2h - h^2)}{\pi^2 (2h - h^2)^2} \\ \\ &= \ \ \lim_{h \to 0} \dfrac{\pi h^2}{\pi^2 h^2 (2 - h)^2} \\ \\ &= \ \ \lim_{h \to 0} \dfrac{1}{\pi (2 - h)^2} \\ \\ &= \ \ \dfrac{1}{4 \pi} \approx \boxed{0.07957747} \end{array}$

There is probably a more elegant solution, but here is mine:

Let $r$ be the horizontal radius of the spherical cap, which itself has a height of $h$ , so the distance from the center of the sphere and the horizontal plane of the spherical cap is $1-h$ . Before proceeding with the solution, first note that the area of the horizontal plane of the spherical cap is $a = \pi r^2$ and the surface area of the spherical cap itself is given by $A = \pi (r^2 + h^2)$ or $A = a + \pi h^2$

The following relationship holds: $r^2 + (1-h)^2 = 1$ Or when expanded: $r^2 + h^2 + 1 - 2h = 1$ Multiplying both sides by $h$ and rearranging a bit yields: $\pi (r^2 + h^2) = 2\pi h$ Notice that the left hand side is precisely $A$ , so: $A = 2 \pi h$ or $h = \frac{A}{2\pi}$ $h^2 = \frac{A^2}{4 \pi^2}$ I squared both sides because now we can plug into the equation $A = a + \pi h^2$ : $A = a + \pi \frac{A^2}{4 \pi^2}$ Rearranging again we get this quadratic equation in terms of A: $A^2 - 4 \pi A + 4 \pi a = 0$ Using quadratic formula or completing the square yields: $(A-2 \pi)^2 = 4 \pi ( \pi - a)$ $A = 2 \pi \pm 2 \sqrt{ \pi ( \pi - a)}$ We get two possible solutions for $A(a)$ but clearly, the one of interest is with the negative sign since, by corollary, if $a \rightarrow 0$ then so must $A$ . Therefore: $A(a) = 2 \pi - 2 \sqrt{ \pi ( \pi - a)}$ Now we found $A$ in terms of $a$ we can evaluate the limit: $\lim_{a\to0} \frac{A(a)-a}{a^2}$ $\lim_{a\to0} \frac{2 \pi - 2 \sqrt{ \pi ( \pi - a)}-a}{a^2}$ From here on, multiple cases of undeterminate forms $\frac{"0"}{0}$ emerge so applying L'Hopital's rule twice will finally yield: $\lim_{a\to0} \frac{\sqrt{\pi}(\pi-a)^{-3/2}}{4}$ $= \frac{1}{4 \pi} \approx 0.0796$

From the above figure we have, $r = R \sin A = \sin A$ since $R = 1$ $\implies$ area of cross-section, $a = \pi*r^2 = \pi\sin^2A$

$h = R - R\cos A = 1 - \cos A$ . Curved surface area of minor cap is, $A(a) =$ $2\pi hR = 2\pi(1 - \cos A)$

Now, $\quad\quad \large \lim_{a \to 0} \frac{A(a) - a}{a^2} = \lim_{A \to 0}\frac{2\pi(1 - \cos A) - \pi\sin^2A}{(\pi\sin^2A)^2}\space\space$ [since $\large a = \pi\sin^2A$ , $\large a\to0 \implies A\to 0]$

$=\large \frac{1}{\pi}\times\lim_{A \to 0} \frac{4\sin^2\big(\frac{A}{2}\big) - 4\sin^2\big(\frac{A}{2}\big)\cos^2\big(\frac{A}{2}\big)}{\sin^4A} = \large \frac{1}{\pi}\times\lim_{A \to 0} \frac{4\sin^2\big(\frac{A}{2}\big)\bigg(1 - \cos^2\big(\frac{A}{2}\big)\bigg)}{\sin^4A}$

$= \LARGE \frac{4}{\pi}\times\lim_{A \to 0} \frac{\sin^4\big(\frac{A}{2}\big)}{\sin^4A} = \frac{4}{\pi}\times\lim_{A \to 0} \frac{A^4 \sin^4\big(\frac{A}{2}\big)}{2^4\frac{A^4}{2^4}\sin^4A} = \frac{4}{\pi}\times\frac{1}{2^4} = \frac{1}{4\pi} \approx \boxed{0.0795}$