In a $∆$ABC, angles A, B, C are in A.P. then $\underset{A\to C}{lim} \frac{\sqrt{3-4\sin A\sin C}}{|A-C|}$

A, B, C are in A.P. $\Rightarrow$ B = 60°

$\begin{array}{l}\Rightarrow \cos B=\cos {60}^{\circ }=\frac{1}{2}=\frac{c^2+a^2-b^2}{2ca}\\ \Rightarrow a^2+c^2=b^2+ac\\ \Rightarrow (a-c{)}^2=b^2-ac\\ \Rightarrow |a-c|=\sqrt{b^2-ac}\\ \Rightarrow |\sin A-\sin C|=\sqrt{{\sin }^2 B-\sin A\sin C}\end{array}$

$\begin{array}{l}\Rightarrow 2\cos \frac{A+C}{2}\left|\sin \frac{A-C}{2}\right|=\sqrt{\frac{3}{4}-\sin A\sin C}\\ \Rightarrow 2 \frac{1}{2}\left|\sin \frac{A-C}{2}\right|=\sqrt{\frac{3-4\sin A\sin c}{4}}\end{array}$

so that $\underset{A\to C}{lim} \frac{\sqrt{3-4\sin A\sin C}}{|A-C|}=\underset{A\to C}{lim} \frac{2\sin \left|\frac{A-C}{2}\right|}{|A-C|}=1$.