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!
 A: \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*}
A: 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 $\frac 32 \sin 2a$
A: 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
$$\frac 32 \sin(2a)+\frac{1}{4} (9 \cos (2 a)-1)\,x+O\left(x^2\right)$$
