QJEEM20Maa8ed: JEE Main 2020 Mathematics - Class 12 JEEM20Maa8ed Solution

Question

Verified Solution • 2025–26 Session moderate

\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}

is equal to

Accepted Answer

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}}}

Step-by-Step Verified Solution