The value oflimx→0 sin2⁡π2−axsec2⁡π2−bxis equal to

1∞ form ⇒elimx→0sin2π2-ax-1.sec2π2-bx=elimx→0-cos2π2-axcos2π2-bx By L-rule=elimx→0-2 sinπ2-axcosπ2-ax.-πa2-ax22sinπ2-bxcos π2-bx.-πb2-bx2 =elimx→0-2 sinπ2cosπ2-ax.-πa222sinπ2cos π2-bx.-πb22=elimx→0- sinπ2-ax-πa2-ax2.asinπ2-bx-πb2-bx2.b =e-a2b2⇒K=1

The value oflimx→0 sin2⁡π2−axsec2⁡π2−bxis equal to

1∞ form ⇒elimx→0sin2π2-ax-1.sec2π2-bx=elimx→0-cos2π2-axcos2π2-bx By L-rule=elimx→0-2 sinπ2-axcosπ2-ax.-πa2-ax22sinπ2-bxcos π2-bx.-πb2-bx2 =elimx→0-2 sinπ2cosπ2-ax.-πa222sinπ2cos π2-bx.-πb22=elimx→0- sinπ2-ax-πa2-ax2.asinπ2-bx-πb2-bx2.b =e-a2b2⇒K=1

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

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$e^{- a / b}$

$e^{- a^{2} / b^{2}}$

$a^{2 a / b}$

$a^{4 a / b}$

answer is B.

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Detailed Solution

$1^{\infty} form \Rightarrow e^{\lim_{x \to 0} (\sin^{2} \frac{π}{2 - ax} - 1) . \sec^{2} \frac{π}{2 - bx}} = e^{\lim_{x \to 0} \frac{- \cos^{2} (\frac{π}{2 - ax})}{\cos^{2} (\frac{π}{2 - bx})}} By L - rule = e^{\lim_{x \to 0} \frac{- 2 \sin (\frac{π}{2 - ax}) \cos (\frac{π}{2 - ax}) . \frac{- πa}{{(2 - ax)}^{2}}}{2 \sin (\frac{π}{2 - bx}) \cos \frac{π}{2 - bx} . \frac{- πb}{{(2 - bx)}^{2}}}} = e^{\lim_{x \to 0} \frac{- 2 \sin (\frac{π}{2}) \cos (\frac{π}{2 - ax}) . \frac{- πa}{{(2)}^{2}}}{2 \sin (\frac{π}{2}) \cos \frac{π}{2 - bx} . \frac{- πb}{{(2)}^{2}}}} = e^{\lim_{x \to 0} \frac{- \sin (\frac{π}{2 - ax}) \frac{- πa}{{(2 - ax)}^{2}} . a}{\sin (\frac{π}{2 - bx}) \frac{- πb}{{(2 - bx)}^{2}} . b}} = e^{\frac{- a^{2}}{b^{2}}} \Rightarrow K = 1$