Question

Find b

$\underset{x\to 0}{lim}\left[\frac{\sin (x+a)+\sin (a-x)-2\sin a}{x\sin x}\right]$ $=-b\sin a.$

Answer 1

$\underset{x\to 0}{lim}\left[\frac{\sin (x+a)+\sin (a-x)-2\sin a}{x\sin x}\right].$

$=\underset{x\to 0}{lim}\left[\frac{2\sin a.\cos x-2\sin a}{x\sin x}\right]$

$=\underset{x\to 0}{lim}\left[\frac{2\sin a(\cos x-1)}{x\sin x}\right].$

$=-2\sin a\underset{x\to 0}{lim}\left[\frac{2{\sin }^2\frac{x}{2}}{2x\sin \frac{x}{2}\cos \frac{x}{2}}\right]$

$=-2\sin a\underset{x\to 0}{lim}\left[\frac{\tan \frac{x}{2}}{x}\right]$

$=-\sin a$

$\Rightarrow -\sin a=-bsina$

$\Rightarrow b=1$