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

$$\lim_{x\to0}\frac{\sin(x^2+x)}{x}=1$$
$$A=\lim_{x\to0}\frac{\sin(x^2+x)}{x}$$ $$A=\lim_{x\to0}\frac{\sin(x^2+x)}{x^2+x}\times\lim_{x\to0}\frac{x^2+x}{x}$$ For $x\ne0$ and $x\ne-1$, $x^2+x\ne0$. So, $\lim_{x\to0}\frac{\sin(x^2+x)}{x^2+x}=1$. $$A=1\times\lim_{x\to0}\frac{x^2+x}{x}$$ $$A=\lim_{x\to0}(x+1)$$ $$A=0+1=1$$