Question

The ${lim}_{y\to 0}\{\left(\sin \frac{y-a}{2}\right).\left(\tan \frac{\pi y}{2a}\right)\}$ is

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

Answer 1

Answer: (C)

$-\frac{a}{\pi }$

Let $L=\underset{y\to a}{lim}\left\{\left(\sin \frac{y-a}{2}\right)\left(\tan \frac{\pi y}{2a}\right)\right\}$
$=\underset{y\to a}{lim}\frac{\sin \frac{y-a}{2}}{\cos \frac{\pi y}{2a}}$ ... $[\because \sin \frac{\pi }{2}=1]$
Using by L's Hospital rule we get
$=\underset{y\to a}{lim}\frac{\frac{1}{2}\cos \frac{y-a}{2}}{-\frac{\pi }{2a}{\text{cosec}}^2\frac{\pi y}{2a}}=\frac{\frac{1}{2}\times 1}{-\frac{\pi }{2a}.1}=\frac{-a}{\pi }$