Question

$\underset{x\to 0}{\lim }\frac{\sin \left(3+x\right)-\sin \left(3-x\right)}{x}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{\sin \left(3+x\right)-\sin \left(3-x\right)}{x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2\cos \left({\displaystyle \frac{3+x+3-x}{2}}\right)\sin \left({\displaystyle \frac{3+x-3+x}{2}}\right)}{x}\right]\left\{\because \sin \mathrm{C}-\sin \mathrm{D}=2\cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right)\sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\right\} \\ =\underset{x\to 0}{\lim }\left[\frac{2\cos 3\sin x}{x}\right] \\ =2\cos 3 \end{gathered}$$