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\(\rm \lim_{x\rightarrow{\pi\over2}}{\tan2x\over{\pi\over2}-x}\)
1. -1
2. -2
3. -3
4. -4

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Correct Answer - Option 2 : -2


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.


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

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