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