Correct choice is (b) e^-0.5
Easiest explanation: y=\(\lim_{x\rightarrow 0}(sin(2x))^{tan^2(2x)}\)
Taking log of both side
\(ln y=\lim_{x\rightarrow 0}\frac{ln(sin(2x))}{cot^2(2x)}(0/0)\)
By L’Hospital Rule
\(ln y=-\lim_{x\rightarrow 0}\frac{2cos(2x)}{sin(2x).4.cosec^2(2x)cot(2x)}=-0.5\lim_{x\rightarrow 0}sin^2(2x)\)=-0.5
=>y=e^-0.5