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Mathematics · Limits, Continuity and Differentiability

JEE Advanced 2020 — Paper 1 — Question 45

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

lim⁡x→0+(1−x)1/x−e−1xa\lim _{x \rightarrow 0^{+}} \frac{(1-x)^{1 / x}-e^{-1}}{x^{a}} is equal to a nonzero real number, is ____\_\_\_\_

Answer: 1

Numerical answer — enter this value.

Step-by-step solution

lim⁡x→0+(1−x)1/x−1exa=lim⁡x→0+eln⁡(1−x)x−1exa\lim _{x \rightarrow 0^{+}} \frac{(1-x)^{1 / x}-\frac{1}{e}}{x^{a}}=\lim _{x \rightarrow 0^{+}} \frac{e^{\frac{\ln (1-x)}{x}}-\frac{1}{e}}{x^{a}}

=lim⁡x→0+1e(e−x−x22x+1−1)xa=lim⁡x→01e(e−x2−1xa)=\lim _{x \rightarrow 0^{+}} \frac{1}{e} \frac{\left(e^{\frac{-x-\frac{x^{2}}{2}}{x}+1}-1\right)}{x^{a}}=\lim _{x \rightarrow 0} \frac{1}{e}\left(\frac{e^{-\frac{x}{2}}-1}{x^{a}}\right)

For value of limit to be a non-zero real number a=1\mathrm{a}=1

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 1
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Evaluation of Limit of Functions

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