JEE Advanced 2020 Paper 1Question 1
Let $e$ denote the base of the natural logarithm. The value of the real number $a$ for which the right hand limit $$\lim_{x\to0^+}\frac{(1-x)^{\frac1x}-e^{-1}}{x^a}$$ is equal to a nonzero real number, is ______
Limits & Continuity
Answer$1$
Solution
Concept. Expand $(1-x)^{1/x}$ around $e^{-1}$; the first non-zero term of the numerator is of order $x$. Formulas used. (1) $\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\dots$(2) $e^{u}=1+u+O(u^2)$Step 1 - Logarithm (1). $$\frac1x\ln(1-x)=-1-\frac x2-\frac{x^2}{3}-\dots$$ Step 2 - Exponentiate (2). $$(1-x)^{1/x}=e^{-1}\cdot e^{-\frac x2-\dots}=e^{-1}\Big(1-\frac x2+O(x^2)\Big)$$ Step 3 - Numerator. $(1-x)^{1/x}-e^{-1}=-\frac{x}{2e}+O(x^2)$. Step 4 - Choose $a$. $$\frac{-\frac{x}{2e}+O(x^2)}{x^a}=x^{1-a}\Big(-\frac{1}{2e}+O(x)\Big)$$ If $a<1$ this $\to0$; if $a>1$ it $\to-\infty$. Only $a=1$ gives a non-zero finite limit, $-\frac{1}{2e}$. Answer: 1