JEE Advance - Mathematics (2020 - Paper 1 Offline - No. 18)
let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit
$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{(1 - x)}^{1/x}} - {e^{ - 1}}} \over {{x^a}}}$$
is equal to a non-zero real number, is .............
$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{(1 - x)}^{1/x}} - {e^{ - 1}}} \over {{x^a}}}$$
is equal to a non-zero real number, is .............
Answer
1
Explanation
The right hand limit
$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{(1 - x)}^{1/x}} - {e^{ - 1}}} \over {{x^a}}}$$
$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{e^{\left\{ {{1 \over x}{{\log }_e}(1 - x)} \right\}}} - {e^{ - 1}}} \over {{x^a}}}$$
$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{e^{{1 \over x}\left( { - x - {{{x^2}} \over 2} - {{{x^3}} \over 3} - ...} \right)}} - {e^{ - 1}}} \over {{x^a}}}$$
$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{e^{ - 1}}.{e^{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ....} \right)}} - {e^{ - 1}}} \over {{x^a}}}$$
$$ = {e^{ - 1}}\mathop {\lim }\limits_{x \to {0^ + }} {{{e^{ \left( {-{x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ....} \right)}} - 1} \over {{x^a}}}$$
$$ = {e^{ - 1}}\mathop {\lim }\limits_{x \to {0^ + }} {{{e^{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ...} \right)}} - 1} \over {^{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ...} \right)}}} \times {{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ...} \right)} \over {{x^a}}}$$
$$ = {e^{ - 1}}\mathop {\lim }\limits_{x \to {0^ + }} {{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ...} \right)} \over {{x^a}}}$$
$$ = {e^{ - 1}}\mathop {\lim }\limits_{x \to {0^ + }} \left( { - {1 \over 2}{x^{1 - a}} - {1 \over 3}{x^{2 - a}} - ...} \right)$$
The above limit will be non-zero, if a = 1. And at a = 1, the value of the limit is
$$ = {e^{ - 1}}\left( { - {1 \over 2}} \right) = - {1 \over {2e}}$$
$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{(1 - x)}^{1/x}} - {e^{ - 1}}} \over {{x^a}}}$$
$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{e^{\left\{ {{1 \over x}{{\log }_e}(1 - x)} \right\}}} - {e^{ - 1}}} \over {{x^a}}}$$
$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{e^{{1 \over x}\left( { - x - {{{x^2}} \over 2} - {{{x^3}} \over 3} - ...} \right)}} - {e^{ - 1}}} \over {{x^a}}}$$
$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{e^{ - 1}}.{e^{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ....} \right)}} - {e^{ - 1}}} \over {{x^a}}}$$
$$ = {e^{ - 1}}\mathop {\lim }\limits_{x \to {0^ + }} {{{e^{ \left( {-{x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ....} \right)}} - 1} \over {{x^a}}}$$
$$ = {e^{ - 1}}\mathop {\lim }\limits_{x \to {0^ + }} {{{e^{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ...} \right)}} - 1} \over {^{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ...} \right)}}} \times {{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ...} \right)} \over {{x^a}}}$$
$$ = {e^{ - 1}}\mathop {\lim }\limits_{x \to {0^ + }} {{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ...} \right)} \over {{x^a}}}$$
$$ = {e^{ - 1}}\mathop {\lim }\limits_{x \to {0^ + }} \left( { - {1 \over 2}{x^{1 - a}} - {1 \over 3}{x^{2 - a}} - ...} \right)$$
The above limit will be non-zero, if a = 1. And at a = 1, the value of the limit is
$$ = {e^{ - 1}}\left( { - {1 \over 2}} \right) = - {1 \over {2e}}$$
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