Answer
$$
\lim _{x \rightarrow 0} \frac{\sin (a+2 x)-2 \sin (a+x)+\sin a}{x^{2}} =-\sin a
$$
Work Step by Step
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sin (a+2 x)-2 \sin (a+x)+\sin a}{x^{2}} = \\
&=\lim _{x \rightarrow 0} \frac{\sin a \cos 2 x+\cos a \sin 2 x-2 \sin a \cos x-2 \cos a \sin x+\sin a}{x^{2}} \\
&=\lim _{x \rightarrow 0} \frac{\sin a(\cos 2 x-2 \cos x+1)+\cos a(\sin 2 x-2 \sin x)}{x^{2}} \\ &=\lim _{x \rightarrow 0} \frac{\sin a\left(2 \cos ^{2} x-1-2 \cos x+1\right)+\cos a(2 \sin x \cos x-2 \sin x)}{x^{2}} \\ &=\lim _{x \rightarrow 0} \frac{2(\cos x-1)[\sin a \cos x+\cos a \sin x)(\cos x+1)}{x^{2}(\cos x+1)} \\ &=\lim _{x \rightarrow 0} \frac{-2 \sin ^{2} x[\sin (a+x)]}{x^{2}(\cos x+1)}\\
&=-2 \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{2} \cdot \frac{\sin (a+x)}{\cos x+1}\\
&=-\sin a \end{aligned}
$$
