Correct Answer - Option 2 : -2

__Concept:__

L'Hospital's Rule:

If the limit becomes \({0\over 0}\) or \({\pm\infty\over \pm\infty}\), it is solved by differentiating numerator and denominator.

__Calculation:__

Let L = \(\rm \lim_{x\rightarrow{\pi\over2}}{\tan2x\over{\pi\over2}-x}\)

On putting the limits we get

L = \(\rm \lim_{x\rightarrow{\pi\over2}}{\tan2x\over{\pi\over2}-x}\) = \(0\over0\)

Applying L'Hospital Rule

L = \(\rm \lim_{x\rightarrow{\pi\over2}}{2\sec^2 2x\over-1}\)

Putting the limits we get,

L = \(\rm{-2\sec^2 \pi}\)

L = -2(-1)^{2} =** -2**