JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 25)
If $$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3},$$ then 'a' is equal to :
2
$${3 \over 2}$$
$${2 \over 3}$$
$${1 \over 2}$$
Explanation
Given,
$$\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3}$$
So, $$\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}}\,\left[ {{1^ \propto }\,\,\,form} \right]$$
$$ = {e^{\mathop {\lim }\limits_{x \to \propto } \left[ {\left( {1 + {a \over x} - {4 \over {{x^2}}} - 1} \right)2x} \right]}}$$
$$ = {e^{\mathop {\lim }\limits_{x \to \propto } \left( {2a - {8 \over x}} \right)}}$$
$$ = {e^{2a}}$$
$$ \therefore $$ e2a = e3
$$ \therefore $$ 2a = 3
$$ \Rightarrow $$ a $$=$$ $${3 \over 2}$$
$$\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3}$$
So, $$\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}}\,\left[ {{1^ \propto }\,\,\,form} \right]$$
$$ = {e^{\mathop {\lim }\limits_{x \to \propto } \left[ {\left( {1 + {a \over x} - {4 \over {{x^2}}} - 1} \right)2x} \right]}}$$
$$ = {e^{\mathop {\lim }\limits_{x \to \propto } \left( {2a - {8 \over x}} \right)}}$$
$$ = {e^{2a}}$$
$$ \therefore $$ e2a = e3
$$ \therefore $$ 2a = 3
$$ \Rightarrow $$ a $$=$$ $${3 \over 2}$$
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