Question

Solve $\underset{x\to 1}{lim}\left[\sec \left(\frac{\pi x}{2}\right)\log x\right]$

• A. $-\frac{2}{\pi }$
• B. $-\frac{\pi }{2}$
• C. $-\frac{1}{\pi }$
• D. $\frac{2}{\pi }$

Answer 1

Answer: (A)

$-\frac{2}{\pi }$

$\underset{x\to 1}{lim}\left[\sec \left(\frac{\pi x}{2}\right)logx\right]$

$x-1=t$

$x=1+t$

$t\to 0$

$\underset{t\to 0}{lim}\left[\sec \left(\frac{\pi }{2}\right(1+t\left)log\right(1+t\left)\right)\right]$

$\underset{t\to 0}{lim}(-cosec\left(\frac{\pi }{2}t\right)log(1+t\left)\right)$

$-\underset{t\to 0}{lim}\left(\frac{log(1+t)}{\sin \left(\frac{n}{2}t\right)}\right)$

$=-\underset{t\to 0}{lim}\frac{log(1+t)}{t}\times \frac{t}{\sin \left(\frac{\pi }{2}\right)t}$

$-1\times \underset{t\to 0}{lim}\frac{\frac{\pi }{2}t}{\sin \frac{\pi }{2}t}\times \frac{2}{\pi }$

$-\frac{2}{\pi }\underset{t\to 0}{lim}\frac{1}{\left(\frac{\sin \frac{\pi }{2}t}{\frac{\pi }{2}}\right)}-1$

$=\frac{-2}{\pi }$.