Answer
$$\frac{6}{7}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{2{{\tan }^{ - 1}}3{x^2}}}{{7{x^2}}} \cr
& {\text{evaluating the limit, we get:}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{2{{\tan }^{ - 1}}3{x^2}}}{{7{x^2}}} = \frac{{2{{\tan }^{ - 1}}3{{\left( 0 \right)}^2}}}{{7{{\left( 0 \right)}^2}}} \cr
& = \frac{0}{0} \cr
& {\text{The limit is }}\frac{0}{0}{\text{, so}}{\text{ we can apply l'Hopital's Rule}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{d/dx\left( {2{{\tan }^{ - 1}}3{x^2}} \right)}}{{d/dx\left( {7{x^2}} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{2\left( {6x} \right)}}{{1 + {{\left( {3{x^2}} \right)}^2}}}}}{{14x}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{12x}}{{14x\left( {1 + 9{x^4}} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{6}{{7\left( {1 + 9{x^4}} \right)}} \cr
& {\text{evaluating the limit, we get:}} \cr
& = \frac{6}{{7\left( {1 + 9{{\left( 0 \right)}^4}} \right)}} = \frac{6}{7} \cr
& {\text{Then}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{2{{\tan }^{ - 1}}3{x^2}}}{{7{x^2}}} = \frac{6}{7} \cr} $$
