Limit question concerning a triangle and its circumcircle $ABC$ is an isosceles triangle ($|AB|=|BC|$). Let $s$ be the length of the altitude from vertex $B$ to side $AC$, and let $m=|AC|$. Given that the radius of the circumcircle of $ABC$ is $2\,\text{cm}$, and the area of the triangle is $k$, calculate: $$ \displaystyle\lim_{s \to 0} \frac{s^2}{k\cdot m} $$ * * * Obviously, $k=s\cdot m / 2 $. How do I proceed thereafter?

With the observation that $k = sm/2$, we want $\lim \frac{2s}{m^2}$.

**Hint:** Express $s$ in terms of $m$.
By power of a point about the midpoint of $AC$, $ s(4-s) = (m/2) (m/2)$. Hence $ s = \frac{1}{2} \left(4 - \sqrt{16- m^2 }\right) $.

**Hint:** Evaluate $\lim \frac{ 4 - \sqrt{16 - m^2 } } { m^2}$.
There are several ways to do so, like:
1) multiply by the conjugate
2) apply L'hopital's rule.