Question

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

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{\sin \left(a+x\right)+\sin \left(a-x\right)-2\sin a}{x\sin x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2\sin \left({\displaystyle \frac{a+x+a-x}{2}}\right)\cos \left({\displaystyle \frac{a+x-a+x}{2}}\right)-2\sin a}{x\sin x}\right]\left\{\because \sin \mathrm{C}+\sin \mathrm{D}=2\sin \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right)\cos \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\right\} \\ =\underset{x\to 0}{\lim }\left[\frac{2\sin a\cos x-2\sin a}{x\sin x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2\sin a\left(\cos x-1\right)}{x\sin x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2\sin a\left(1-2{\sin }^2{\displaystyle \frac{x}{2}}-1\right)}{x\sin x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2\sin a\left(-2{\sin }^2{\displaystyle \frac{x}{2}}\right)}{x\sin x}\right] \\ =-4\sin a\underset{x\to 0}{\lim }\left[\frac{1}{{\displaystyle \frac{x\sin x}{x^2}}}\times \frac{\sin \left({\displaystyle \frac{x}{2}}\right)}{{\displaystyle \frac{x}{2}}}\times \frac{\sin \left({\displaystyle \frac{x}{2}}\right)}{{\displaystyle \frac{x}{2}}}\times \frac{1}{4}\right] \\ =-4\sin a\times \frac{1}{1}\times 1\times 1\times \frac{1}{4} \\ =-\sin a \end{gathered}$$