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
