Let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit $$\lim_{x \rightarrow 0^{+}} \frac{(1-x)^{\frac{1}{x}}-e^{-1}}{x^{a}}$$ is equal to a nonzero real number, is ____.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Limits XII

Let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit

x \to 0^{+} lim \frac{( 1 - x ) ^{\frac{1}{x}} - e ^{- 1}}{x ^{a}}

is equal to a nonzero real number, is ____.

Answer: 1

We have:

x \to 0^{+} lim \frac{( 1 - x ) ^{\frac{1}{x}} - e ^{- 1}}{x ^{a}} = x \to 0^{+} lim \frac{e ^{\frac{l n ( 1 - x )}{x}} - e ^{- 1}}{x ^{a}}

= \frac{1}{e} x \to 0^{+} lim \frac{e ^{\frac{l n ( 1 - x )}{x} + 1} - 1}{x ^{a}}

= \frac{1}{e} x \to 0^{+} lim \frac{\frac{l n ( 1 - x )}{x} + 1}{x ^{a}}

(using

e^{t} - 1 \sim t

t \to 0

)

= \frac{1}{e} x \to 0^{+} lim \frac{ln ( 1 - x ) + x}{x ^{a + 1}}

Using

ln (1 - x) = - x - \frac{x ^{2}}{2} - \frac{x ^{3}}{3} - \dots

= \frac{1}{e} x \to 0^{+} lim \frac{- x - \frac{x ^{2}}{2} - \frac{x ^{3}}{3} - \dots + x}{x ^{a + 1}} = \frac{1}{e} x \to 0^{+} lim \frac{- \frac{x ^{2}}{2} - \frac{x ^{3}}{3} - \dots}{x ^{a + 1}}

For this limit to be a nonzero finite number, we need

a + 1 = 2

, so

a = 1

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Numerical answer