<s>[INST] I need the steps to solving this limit without using l´Hopital rule I've tried many ways of solving this limit without using l'Hopital and I just can't figure it out. I know the answer is $3/2 \sin (2a).$
$$\lim_{x\,\to\,0} \frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x$$
Thank you!
 [/INST] \begin{align*}
&\dfrac{\sin(a+x)\sin(a+2x)-\sin^{2}a}{x}\\
&=2\cdot\dfrac{\sin(a+x)[\sin(a+2x)-\sin a]}{2x}+\dfrac{(\sin a)[\sin(a+x)-\sin a]}{x}\\
&\rightarrow 2\sin(a+0)(\sin x)'\bigg|_{x=a}+(\sin a)(\sin x)'\bigg|_{x=a}\\
&=2(\sin a)(\cos a)+(\sin a)(\cos a).
\end{align*}
The true without L'Hopital:
\begin{align*}
&\dfrac{\sin(a+x)\sin(a+2x)-\sin^{2}a}{x}\\
&=\dfrac{(\sin a\cos x+\cos a\sin x)(\sin a\cos 2x+\cos a\sin 2x)-\sin^{2}a}{x}\\
&=\dfrac{\sin^{2}a\cos x\cos 2x+\cos^{2}a\sin x\sin 2x+\sin a\cos a(\sin 2x\cos x+\cos x\sin 2x)-\sin^{2}a}{x}\\
&=\dfrac{\sin^{2}a(\cos x\cos 2x-1)}{x}+\cos^{2}a\cdot\dfrac{\sin x}{x}\cdot\sin 2x+\dfrac{\sin 2a}{2}\cdot 3\cdot\dfrac{\sin 3x}{3x},
\end{align*}
where
\begin{align*}
\dfrac{\cos x\cos 2x-1}{x}&=\dfrac{\cos x-1}{x}-2\cdot\dfrac{\sin x}{x}\cdot\cos x\\
&=\dfrac{-2\sin^{2}(x/2)}{x}-2\cdot\dfrac{\sin x}{x}\cdot\cos x\\
&\rightarrow 0
\end{align*}
</s>