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

Given limit is of the form 1∞∴Given limit=elimx→0 sec2⁡π2−bxsin2⁡π2−cx−1=e-limx→0 cos2⁡π2−axcos2⁡π2−bxUse L-Hospital rule =e-limx→0sin⁡π2−axsin⁡π2−bxab Again use L-Hospital rule =e−a2b2

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

Given limit is of the form 1∞∴Given limit=elimx→0 sec2⁡π2−bxsin2⁡π2−cx−1=e-limx→0 cos2⁡π2−axcos2⁡π2−bxUse L-Hospital rule =e-limx→0sin⁡π2−axsin⁡π2−bxab Again use L-Hospital rule =e−a2b2

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

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

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

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

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

answer is B.

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Detailed Solution

Given limit is of the form $1^{\infty}$
$∴$ Given $lim it = e^{lim_{x \to 0} \sec^{2} (\frac{π}{2 - bx}) (\sin^{2} (\frac{π}{2 - cx}) - 1)}$
$= e^{- lim_{x \to 0} \frac{\cos^{2} (\frac{π}{2 - ax})}{\cos^{2} (\frac{π}{2 - bx})}}$
Use L-Hospital rule $= e^{- lim_{x \to 0} \frac{\sin (\frac{π}{2 - ax})}{\sin (\frac{π}{2 - bx})} \frac{a}{b}}$
Again use L-Hospital rule $= e^{\frac{- a^{2}}{b^{2}}}$