# Question #389da

If the problem is ${\lim}_{x \to 0} \sin \frac{2 x}{2 {x}^{2} + x}$ then direct substitution gives $\frac{0}{0}$ so we can use L'Hopital's Rule:
${\lim}_{x \to 0} \sin \frac{2 x}{2 {x}^{2} + x} = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \sin \left(2 x\right)}{\frac{d}{\mathrm{dx}} \left(2 {x}^{2} + x\right)}$
$= {\lim}_{x \to 0} \frac{2 \cos \left(2 x\right)}{4 x + 1} = \frac{2}{1} = 2$