## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to0}\frac{x^2-x+\sin x}{2x}=0$$
$$A=\lim_{x\to0}\frac{x^2-x+\sin x}{2x}$$ $$A=\lim_{x\to0}\frac{x^2-x}{2x}+\lim_{x\to0}\frac{\sin x}{2x}=X+Y$$ Consider $X$: $$X=\lim_{x\to0}\frac{x^2-x}{2x}=\lim_{x\to0}\frac{x-1}{2}$$ $$X=\frac{0-1}{2}=-\frac{1}{2}$$ Consider $Y$: $$Y=\lim_{x\to0}\frac{\sin x}{2x}$$ Multiply both numerator and denominator by $1/2$: $$Y=\lim_{x\to0}\frac{1/2\sin x}{1/2\times2x}=\frac{1}{2}\lim_{x\to0}\frac{\sin x}{x}$$ Apply Theorem 7 with $\theta=x$: $$Y=\frac{1}{2}\times1=\frac{1}{2}$$ Therefore, $$A=X+Y=-\frac{1}{2}+\frac{1}{2}=0$$