Question

Solve:
$\underset{x\to 0}{lim}\frac{\tan 2\left(\frac{\pi }{2}.x\right)}{\left(\frac{\pi }{2}-x\right)-\frac{\pi }{2}}$

Answer 1

$\underset{x\to 0}{lim}\frac{\tan 2\left(\frac{\pi }{2}x\right)}{(\frac{\pi }{2}-x-\frac{\pi }{2})}$

Multiply numerator by $\frac{\pi x}{\pi x}$ to convert in the form of $\frac{\tan \theta }{\theta }$ [Where $\theta =\pi x$]
$\Rightarrow \underset{x\to 0}{lim}\frac{\tan \left(\pi x\right)}{(\frac{\pi }{2}-x-\frac{\pi }{2})}=\underset{x\to 0}{lim}\left[\frac{\frac{\tan \left(\pi x\right)}{\pi x}\times \pi x}{(-x)}\right]$

$=\underset{x\to 0}{lim}\left[\frac{\tan \left(\pi x\right)}{\pi x}\right]\times \left[\frac{\pi x}{-x}\right]$

$=-\pi \underset{x\to 0}{lim}\left[\frac{\tan \left(\pi x\right)}{\pi x}\right]$

$=-\pi \underset{\pi x\to 0}{lim}\left[\frac{\tan \left(\pi x\right)}{\pi x}\right]$ [Since, $\underset{\theta \to 0}{lim}\frac{\tan \theta }{\theta }=1$]

$=-\pi \left(1\right)=-\pi$

$\therefore \underset{x\to 0}{lim}\left[\frac{\tan 2\left(\frac{\pi }{2}x\right)}{(\frac{\pi }{2}-x)-\frac{\pi }{2}}\right]=-\pi$