The value of limx→0 sin2 π2-axsec2 π2-bx is

Hint 1∞formlimx→0 sin2 π2-axsec2 π2-bx=elimx→0sin2 π2-ax-1sec2 π2-bx =elimx→0-cos2 π2-axcos2 π2-bx=e-limx→0cos π2-axcos π2-bx2 =e-limx→0sinπ2-axπ2-ax-2-asinπ2-bxπ2-bx-2-b2=e-a2b2

The value of limx→0 sin2 π2-axsec2 π2-bx is

Hint 1∞formlimx→0 sin2 π2-axsec2 π2-bx=elimx→0sin2 π2-ax-1sec2 π2-bx =elimx→0-cos2 π2-axcos2 π2-bx=e-limx→0cos π2-axcos π2-bx2 =e-limx→0sinπ2-axπ2-ax-2-asinπ2-bxπ2-bx-2-b2=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

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$e^{2 a / b}$

$e^{aa / b}$

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

$e^{- a / b}$

answer is B.

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Detailed Solution

$Hint 1^{\infty} form$

$\lim_{x \to 0} {[\sin^{2} (\frac{π}{2 - a x})]}^{\sec^{2} [\frac{π}{2 - b x}]} = e^{\lim_{x \to 0} (\sin^{2} (\frac{π}{2 - a x}) - 1) \sec^{2} [\frac{π}{2 - b x}]} = e^{\lim_{x \to 0} \frac{(- \cos^{2} (\frac{π}{2 - a x}))}{\cos^{2} [\frac{π}{2 - b x}]}} = e^{- {(\lim_{x \to 0} \frac{(\cos (\frac{π}{2 - a x}))}{\cos [\frac{π}{2 - b x}]})}^{2}} = e^{- {(\lim_{x \to 0} \frac{\sin (\frac{π}{2 - a x}) π {(2 - ax)}^{- 2} (- a)}{\sin (\frac{π}{2 - b x}) π {(2 - b x)}^{- 2} (- b)})}^{2}} = e^{- \frac{a^{2}}{b^{2}}}$