Calcule, se existir

$$\mathop {\lim }\limits_{x \to 0} \frac{{\operatorname{sen} x}}{x} = 1$$

1. Ora,
   $$\mathop {\lim }\limits_{x \to 0} \frac{{\operatorname{sen} 3x}}{x} = 3 \times \mathop {\lim }\limits_{x \to 0} \frac{{\operatorname{sen} 3x}}{{3x}} = 3 \times \underbrace {\mathop {\lim }\limits_{y \to 0} \frac{{\operatorname{sen} y}}{y}}_1 = 3$$
2. Ora,
   $$\mathop {\lim }\limits_{\theta  \to 0} \frac{\theta }{{\operatorname{sen} \frac{\theta }{2}}} = \mathop {\lim }\limits_{\theta  \to 0} \frac{1}{{\frac{{\operatorname{sen} \frac{\theta }{2}}}{\theta }}} = \frac{1}{{\frac{1}{2} \times \mathop {\lim }\limits_{\theta  \to 0} \frac{{\operatorname{sen} \frac{\theta }{2}}}{{\frac{\theta }{2}}}}} = \frac{1}{{\frac{1}{2} \times \underbrace {\mathop {\lim }\limits_{y \to 0} \frac{{\operatorname{sen} y}}{y}}_{`1}}} = 2$$
3. Ora,
   $$\begin{array}{*{20}{l}}
   {\mathop {\lim }\limits_{x \to 0} \frac{{\operatorname{sen} 2x}}{{\operatorname{sen} 3x}}}& = &{\mathop {\lim }\limits_{x \to 0} \left( {\frac{2}{3} \times \frac{{\operatorname{sen} 2x}}{{2x}} \times \frac{{3x}}{{\operatorname{sen} 3x}}} \right)} \\
   {}& = &{\frac{2}{3} \times \mathop {\lim }\limits_{x \to 0} \frac{{\operatorname{sen} 2x}}{{2x}} \times \frac{1}{{\mathop {\lim }\limits_{x \to 0} \frac{{\operatorname{sen} 3x}}{{3x}}}}} \\
   {}& = &{\frac{2}{3} \times \underbrace {\mathop {\lim }\limits_{y \to 0} \frac{{\operatorname{sen} y}}{y}}_1 \times \frac{1}{{\underbrace {\mathop {\lim }\limits_{z \to 0} \frac{{\operatorname{sen} z}}{z}}_1}}} \\
   {}& = &{\frac{2}{3}}
   \end{array}$$
4. Ora,
   $$\begin{array}{*{20}{l}}
   {\mathop {\lim }\limits_{} \left[ {n\operatorname{sen} \left( {\frac{{2\pi }}{n}} \right)} \right]}& = &{\mathop {\lim }\limits_{n \to  + \infty } \left[ {n\operatorname{sen} \left( {\frac{{2\pi }}{n}} \right)} \right]} \\
   {}& = &{\mathop {\lim }\limits_{y \to {0^ + }} \left[ {\frac{1}{y}\operatorname{sen} \left( {2\pi y} \right)} \right]} \\
   {}& = &{2\pi  \times \mathop {\lim }\limits_{y \to {0^ + }} \frac{{\operatorname{sen} \left( {2\pi y} \right)}}{{2\pi y}}} \\
   {}& = &{2\pi  \times \underbrace {\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\operatorname{sen} x}}{x}}_1} \\
   {}& = &{2\pi }
   \end{array}$$