Answer
$2$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{\tan 2x}}{x} \cr
& = \mathop {\lim }\limits_{x \to 0} \left[ {\tan 2x\left( {\frac{1}{x}} \right)} \right] \cr
& {\text{Use the quotient identity tan}}\theta = \frac{{\sin \theta }}{{\cos \theta }} \cr
& = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin 2x}}{{\cos 2x}}\left( {\frac{1}{x}} \right)} \right] \cr
& = \mathop {\lim }\limits_{x \to 0} \left[ {\left( {\frac{{\sin 2x}}{x}} \right)\left( {\frac{1}{{\cos 2x}}} \right)} \right] \cr
& {\text{Multiplying and dividing by }}2 \cr
& = \mathop {\lim }\limits_{x \to 0} \left[ {\left( {\frac{{\sin 2x}}{{2x}}} \right)\left( {\frac{2}{{\cos 2x}}} \right)} \right] \cr
& {\text{Use the product law of limits}}{\text{, the limit of a product is the }} \cr
& {\text{product of the limits}}{\text{, then}} \cr
& = \left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{2x}}} \right)\mathop {\lim }\limits_{x \to 0} \left( {\frac{2}{{\cos 2x}}} \right) \cr
& {\text{Letting }}\theta = 2x,{\text{ then }}\theta \to 0{\text{ as }}x \to 0,{\text{ so by }}\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1 \cr
& = \overbrace {\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{2x}}}^{\theta = 2x,{\text{ }}\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1}\left[ {\mathop {\lim }\limits_{x \to 0} \left( {\frac{2}{{\cos 2x}}} \right)} \right] \cr
& = \left( 1 \right)\left[ {\frac{2}{{\cos 2\left( 0 \right)}}} \right] \cr
& = \left( 1 \right)\left( {\frac{2}{1}} \right) \cr
& = 2 \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\tan 2x}}{x} = 2 \cr} $$
