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!

You could also have used Talor series around $x=0$ and even got more than the limit itself
$$\sin(a+k x)=\sin (a)+k  \cos (a)\,x-\frac{1}{2}  k^2 \sin (a)\,x^2+O\left(x^3\right)$$ making
$$\sin(a+ x)\sin(a+2 x)=\sin ^2(a)+3  \sin (a) \cos (a)\,x+ \left(2 \cos ^2(a)-\frac{5 }{2}\sin
   ^2(a)\right)\,x^2+O\left(x^3\right)$$
$$\frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x=3  \sin (a) \cos (a)\,+ \left(2 \cos ^2(a)-\frac{5 }{2}\sin
   ^2(a)\right)\,x+O\left(x^2\right)$$ which shows the limit and how it is approached.
If you simplify, you should notice that the end result is just
$$\frac 32 \sin(2a)+\frac{1}{4} (9 \cos (2 a)-1)\,x+O\left(x^2\right)$$
