Let denote the base of the natural logarithm. The value of the real number for which the right hand limit is equal to a non-zero real number, is ________. [2020]

Question

Let $e$ denote the base of the natural logarithm. The value of the real number $a$ for which the right hand limit $\lim_{x \to 0^{+}} \frac{{(1 - x)}^{\frac{1}{x}} - e^{- 1}}{x^{a}}$ is equal to a non-zero real number, is ________. [2020]

Smart Achievers · Accepted Answer

(1)

$\lim_{x \to 0^{+}} \frac{{(1 - x)}^{1 / x} - e^{- 1}}{x^{a}} = 1$ $= \lim_{x \to 0^{+}} \frac{e^{(\frac{l n (1 - x)}{x})} - \frac{1}{e}}{x^{a}}$ $[∵ {(1 - x)}^{1 / x} = e^{\frac{1}{x} \ln (1 - x)}]$

$= \lim_{x \to 0^{+}} \frac{1}{e} \cdot \frac{e^{(1 + \frac{\ln (1 - x)}{x})} - 1}{x^{a}}$ $= \frac{1}{e} \lim_{x \to 0^{+}} \frac{l n (1 - x) + x}{x^{a + 1}}$

$= \frac{1}{e} \lim_{x \to 0^{+}} \frac{(- x - \frac{x^{2}}{2} - \frac{x^{3}}{3} - \dots) + x}{x^{a + 1}}$

$∴$ $a = 1$

Solution

Q.
Let $e$ denote the base of the natural logarithm. The value of the real number $a$ for which the right hand limit $\lim_{x \to 0^{+}} \frac{{(1 - x)}^{\frac{1}{x}} - e^{- 1}}{x^{a}}$ is equal to a non-zero real number, is ________. [2020]

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Solution

Q. Let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit limx→0+(1-x)1x-e-1xa is equal to a non-zero real number, is ________. [2020]

Ans. (1) limx→0+(1-x)1/x-e-1xa=1=limx→0+e(ln(1-x)x)-1exa [∵ (1-x)1/x=e1xln(1-x)] =limx→0+1e·e(1+ln(1-x)x)-1xa=1elimx→0+ln(1-x)+xxa+1 =1elimx→0+(-x-x22-x33-⋯)+xxa+1 ∴ a=1

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Q.
Let $e$ denote the base of the natural logarithm. The value of the real number $a$ for which the right hand limit $\lim_{x \to 0^{+}} \frac{{(1 - x)}^{\frac{1}{x}} - e^{- 1}}{x^{a}}$ is equal to a non-zero real number, is ________. [2020]