Just as Joffysloffy answered, Taylor series are extremely convenient for solving the problem of the limit and even more.
Rewrite, as Michael Hardy suggested, $$A=\dfrac{x-\sin x\cos x}{\sin x-\sin x\cos x} = \dfrac{2x - \sin(2x)}{2\sin x-\sin(2x)}$$ and use the fact that $$\sin(y)=\sum_{i=0}^{\infty}\frac{(-1)^n}{(2n+1)!}y^{2n+1}$$ So, the numerator write $$\frac{4 x^3}{3}-\frac{4 x^5}{15}+\frac{8 x^7}{315}+O\left(x^8\right)$$ and the denominator $$x^3-\frac{x^5}{4}+\frac{x^7}{40}+O\left(x^8\right)$$ So, performing the long division $$A=\frac{4}{3}+\frac{x^2}{15}+\frac{11 x^4}{1260}+O\left(x^5\right)$$ which gives not only the limit but also how it is approached.
If you plot the function and the approximation for $0 \leq x \leq \frac{\pi}{2}$, you will probably be amazed to notice how close are the two curves.
What is interesting (at least to me !) is that, if you had to solve for $x$ the equation $A=\frac{3}{2}$, the approximation limits the problem to a quadratic in $x^2$ and the solution would be $$x=\sqrt{\frac{1}{11} \left(\sqrt{4074}-42\right)}\approx 1.40867$$ while the exact solution is $\approx 1.37873$. Using only the second order approximation would immediately lead to $x=\sqrt{\frac{5}{2}}\approx 1.58114$ which already a good approximation from which any root-finding method could efficiently start.