Question

$\underset{x\to 0}{lim}\frac{{\log }_{\sec \left(\frac{x}{2}\right)}\cos x}{{\log }_{\sec x}\cos \left(\frac{x}{2}\right)}$

• A. $14$
• B. $15$
• C. $16$
• D. $17$

Answer 1

Answer: (C)

$16$

${lim}_{x\to 0}\frac{{\log }_{\sec \left(x/2\right)}\cos x}{{\log }_{\sec x}\cos \left(x/2\right)}={lim}_{x\to 0}\frac{\log \cos x}{\log \sec \left(x/2\right)}\frac{\log \sec x}{\log \cos \left(x/2\right)}={lim}_{x\to 0}\frac{\log \cos x}{-\log \cos \left(x/2\right)}\frac{-\log \cos x}{\log \cos \left(x/2\right)}={lim}_{x\to 0}{\left(\frac{\log \cos x}{\log \cos \left(x/2\right)}\right)}^2={lim}_{x\to 0}{\left(\frac{\log (1+\cos x-1)}{\cos x-1}\frac{\cos \left(x/2\right)-1}{\log (1+\cos (x/2)-1)}\frac{\cos x-1}{\cos \left(x/2\right)-1}\right)}^2$

$={lim}_{x\to 0}{\left(\frac{\cos x-1}{\cos \left(x/2\right)-1}\right)}^2={lim}_{x\to 0}{\left(\frac{\cos 2t-1}{\cos t-1}\right)}^2$ (Where $x=2t$)

$={lim}_{x\to 0}{\left(\frac{{2\sin }^{2}t}{t^2}\frac{4{\left(t/2\right)}^2}{2{\sin }^2\left(t/2\right)}\right)}^2=16$