\(\displaystyle\lim_{x\to \frac{\pi}{2}} \tan^2 x\left(\sqrt{2\sin^2 x+3\sin x+4}-\sqrt{\sin^2 x+6\sin x+2}\right)\) is equal to
$\frac{1}{12}$
\(A=2\sin^2x+3\sin x+4,\ B=\sin^2x+6\sin x+2\), so \(A-B=\sin^2x-3\sin x+2=(\sin x-1)(\sin x-2)\).
Rationalising, \(\sqrt{A}-\sqrt{B}=\dfrac{A-B}{\sqrt{A}+\sqrt{B}}=\dfrac{(\sin x-1)(\sin x-2)}{\sqrt{A}+\sqrt{B}}\).
Also \(\tan^2x=\dfrac{\sin^2x}{\cos^2x}=\dfrac{\sin^2x}{(1-\sin x)(1+\sin x)}\).
Multiplying, and using \((\sin x-1)=-(1-\sin x)\) to cancel the common factor, the expression simplifies to \(\dfrac{\sin^2x(2-\sin x)}{(1+\sin x)(\sqrt{A}+\sqrt{B})}\).
As \(x\to \pi/2\), \(\sin x\to 1\), so \(A\to 9,\ B\to 9\), giving \(\sqrt{A}+\sqrt{B}\to 6\).
The limit equals \(\dfrac{1^2\cdot(2-1)}{(1+1)\cdot 6}=\dfrac{1}{12}\).
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