Official solution
Given \[\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}\]
Putting x = 0 we get 1\[\infty \] form.
\[ \therefore \] \[{e^{\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{{3{x^2} + 2} \over {7{x^2} + 2}} - 1} \right]}}\]
= \[{e^{\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{{ - 4{x^2}} \over {7{x^2} + 2}}} \right]}}\]
= e-4/2
= e-2 = \[{1 \over {{e^2}}}\]