JEE MAIN - Mathematics (2020 - 8th January Morning Slot - No. 16)
$$\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}$$ is equal to
e
e2
$${1 \over {{e^2}}}$$
$${1 \over e}$$
Explanation
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 = $${1 \over {{e^2}}}$$
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 = $${1 \over {{e^2}}}$$
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