The value of limx→0 sin2⁡π2−axsec2⁡π2−bx is equal to

Given limit is of the form 1∞ ∴ Given limitL=limx→0 sin2π2−axsec2π2−bxL=elimx→0 sec2⁡π2−bxlog⁡sin2⁡π2−axL=elimx→0 log1−cos2π2−axcos2π2−bxL=elimx→0 log⁡1−cos2⁡π2−ax⋅−cos2⁡π2−ax−cos2⁡π2−ax⋅cos2⁡π2−bx limx→0 log1+xx=1L=elimx→0 −cos2⁡π2−axcos2⁡π2−bx L=e−limx→0 cos⁡π2−axcos⁡π2−bx2 ⇒ (Use L-Hospital rule) ⇒e-a2b2

The value of limx→0 sin2⁡π2−axsec2⁡π2−bx is equal to

Given limit is of the form 1∞ ∴ Given limitL=limx→0 sin2π2−axsec2π2−bxL=elimx→0 sec2⁡π2−bxlog⁡sin2⁡π2−axL=elimx→0 log1−cos2π2−axcos2π2−bxL=elimx→0 log⁡1−cos2⁡π2−ax⋅−cos2⁡π2−ax−cos2⁡π2−ax⋅cos2⁡π2−bx limx→0 log1+xx=1L=elimx→0 −cos2⁡π2−axcos2⁡π2−bx L=e−limx→0 cos⁡π2−axcos⁡π2−bx2 ⇒ (Use L-Hospital rule) ⇒e-a2b2

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The value of $lim_{x \to 0} {\{\sin^{2} (\frac{π}{2 - a x})\}}^{\sec^{2} \frac{π}{2 - b x}}$ is equal to

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$e^{\frac{4 a}{b}}$

$e^{- \frac{a^{2}}{b^{2}}}$

$e^{- \frac{α}{b}}$

$e^{\frac{2 a}{b}}$

answer is B.

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Detailed Solution

Given limit is of the form $1^{\infty} ∴$ Given limit

$L = lim_{x \to 0} {\{\sin^{2} (\frac{π}{2 - a x})\}}^{\sec^{2} (\frac{π}{2 - b x})}$

$L = e^{(lim_{x \to 0} \sec^{2} (\frac{π}{2 - b x}) (\log \sin^{2} (\frac{π}{2 - a x})))}$

$L = e^{(lim_{x \to 0} \frac{\log (1 - \cos^{2} (\frac{π}{2 - a x}))}{\cos^{2} (\frac{π}{2 - b x})})}$

$L = e^{(lim_{x \to 0} \frac{(\log (1 - \cos^{2} (\frac{π}{2 - a x}))) \cdot (- \cos^{2} (\frac{π}{2 - a x}))}{(- \cos^{2} (\frac{π}{2 - a x})) \cdot \cos^{2} (\frac{π}{2 - b x})})} lim_{x \to 0} \frac{\log (1 + x)}{x} = 1$

$L = e^{(lim_{x \to 0} - \frac{\cos^{2} (\frac{π}{2 - a x})}{\cos^{2} (\frac{π}{2 - b x})})} L = {e^{- (lim_{x \to 0} \frac{\cos (\frac{π}{2 - a x}}{\cos (\frac{π}{2 - b x})})}}^{2}$