\[\boxed{\mathbf{12.}}\]
\[\mathbf{а)\ }\lim_{x \rightarrow 0}\frac{\sin{2x}}{2x} = 1\]
\[\mathbf{б)\ }\lim_{x \rightarrow 0}\frac{{3sin}\frac{x}{3}}{x} = \lim_{x \rightarrow 0}\frac{\sin\frac{x}{3}}{\frac{x}{3}} = 1\]
\[\mathbf{в)\ }\lim_{x \rightarrow 0}\frac{\sin\text{πx}}{\text{πx}} = 1\]