Question

The value of $\underset{x\to 0}{lim}{\left({\sin }^2\left(\frac{\pi }{2-ax}\right)\right)}^{{\sec }^2\left(\frac{\pi }{2-bx}\right)}$ is equal to

• A. $e^{-a/b}$
• B. $e^{-a^2/b^2}$
• C. $e^{2a/b}$
• D. $e^{4a/b}$

Answer 1

The correct option is B $e^{-a^2/b^2}$

$L=\underset{x\to 0}{lim}{\left({\sin }^2\left(\frac{\pi }{2-ax}\right)\right)}^{{\sec }^2\left(\frac{\pi }{2-bx}\right)}$

$L=e^{\left(\underset{x\to 0}{lim}{\sec }^2\left(\frac{\pi }{2-bx}\right)\left(\log {\sin }^2\left(\frac{\pi }{2-ax}\right)\right)\right)}$

$L=e^{\left(\underset{x\to 0}{lim}\frac{\log (1-{\cos }^2\left(\frac{\pi }{2-ax}\right))}{{\cos }^2\left(\frac{\pi }{2-bx}\right)}\right)}$

$\because {lim}_{x\to 0}\frac{\log (1+x)}{x}=1$

$L=e^{\left(\underset{x\to 0}{lim}\frac{\left(\log (1-{\cos }^2\left(\frac{\pi }{2-ax}\right))\right)\cdot (-{\cos }^2\left(\frac{\pi }{2-ax}\right))}{(-{\cos }^2\left(\frac{\pi }{2-ax}\right))\cdot {\cos }^2\left(\frac{\pi }{2-bx}\right)}\right)}$

$L=e^{(\underset{x\to 0}{lim}-\frac{{\cos }^2\left(\frac{\pi }{2-ax}\right)}{{\cos }^2\left(\frac{\pi }{2-bx}\right)})}=e^{-{\left({lim}_{x\to 0}\frac{\cos \left(\frac{\pi }{2-ax}\right)}{\cos \left(\frac{\pi }{2-bx}\right)}\right)}^2}$
Using L'Hospital's rule,

$L=e^{-{\left(\underset{x\to 0}{lim}\frac{\sin \left(\frac{\pi }{2-ax}\right)\cdot (2-ax{)}^{-2}\cdot (-a)}{\sin \left(\frac{\pi }{2-bx}\right)\cdot (2-bx{)}^{-2}\cdot (-b)}\right)}^2}=e^{-a^2/b^2}$