# $\rm \lim_{x\rightarrow{\pi\over2}}{\tan2x\over{\pi\over2}-x}$

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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

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