Answer
$$A=-1$$
Work Step by Step
$$A=\lim_{x\to0}\frac{\sin x}{2x^2-x}$$ $$A=\lim_{x\to0}\frac{\sin x}{x(2x-1)}$$ $$A=\lim_{x\to0}\frac{\sin x}{x}\times\lim_{x\to0}\frac{1}{2x-1}$$
We have $\lim_{x\to0}\frac{\sin x}{x}=1$. Therefore, $$A=1\times\lim_{x\to0}\frac{1}{2x-1}=\lim_{x\to0}\frac{1}{2x-1}$$ $$A=\frac{1}{2\times0-1}$$ $$A=-1$$
