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JEE Advanced 2020 · Paper 1, Mathematics question 18

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

lim⁡x→0+(1−x)1x−e−1xa\lim_{x\to0^+}\frac{(1-x)^{\frac1x} - e^{-1}}{x^a}

is equal to a nonzero real number, is ______

Answer and worked solution

Official answer: 1

Expand.
(1−x)1/x=exp⁡ ⁣(ln⁡(1−x)x)=exp⁡ ⁣(−1−x2−x23−⋯ )=e−1(1−x2+O(x2)).(1-x)^{1/x}=\exp\!\left(\frac{\ln(1-x)}x\right)=\exp\!\left(-1-\frac x2-\frac{x^2}3-\cdots\right)=e^{-1}\left(1-\frac x2+O(x^2)\right).
So the numerator is −x2e+O(x2)-\frac{x}{2e}+O(x^2).

The limit. Dividing by xax^a gives a finite nonzero limit only for a=1a=1, where it equals −12e-\frac1{2e}.

Answer: 1

Topics:Limits
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