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!

Perhaps not the most elegant, but:
$\lim_\limits{x\to0} \frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x\\
\lim_\limits{x\to 0} \frac{\frac 12 (-\cos(2a + 3x) + cos x) - \sin^2(a)} x\\
\lim_\limits{x\to 0} \frac{\frac 12 (-\cos(2a)\cos 3x + \sin(2a)\sin (3x) + \cos x) - \sin^2(a)} x\\
\lim_\limits{x\to 0} \frac{\frac 12 (-\cos(2a)\cos 3x + \sin(2a)\sin (3x) + \cos x) - \frac 12 (1-\cos 2a)} x\\
\lim_\limits{x\to 0} \frac{-\cos(2a)\cos 3x + \sin(2a)\sin (3x) + \cos x - 1 +\cos 2a} {2x}\\
\lim_\limits{x\to 0} \frac{\cos(2a)(1-\cos 3x)}{2x} - \frac {1-\cos x}{2x} + \frac { \sin(2a)\sin (3x)}{2x}$
The first term evaluates to $0,$ the second term evaluates to $0,$ the third term evaluates to $\frac 32 \sin 2a$
