Let e denote the base of the natural logarith | vedclass

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(B) Let $L = \lim_{x \rightarrow 0^{+}} \frac{(1-x)^{\frac{1}{x}}-e^{-1}}{x^a}$.We know that $(1-x)^{\frac{1}{x}} = e^{\frac{1}{x} \ln(1-x)}$.Using th

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(B) Let $L = \lim_{x \rightarrow 0^{+}} \frac{(1-x)^{\frac{1}{x}}-e^{-1}}{x^a}$.We know that $(1-x)^{\frac{1}{x}} = e^{\frac{1}{x} \ln(1-x)}$.Using the Taylor series expansion for $\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots$,we have:$\frac{1}{x} \ln(1-x) = -1 - \frac{x}{2} - \frac{x^2}{3} - \dots$Thus,$(1-x)^{\frac{1}{x}} = e^{-1 - \frac{x}{2} - \frac{x^2}{3} - \dots} = e^{-1} \cdot e^{-\frac{x}{2} - \frac{x^2}{3} - \dots}$.Using $e^u = 1 + u + \frac{u^2}{2!} + \dots$,we get:$(1-x)^{\frac{1}{x}} = e^{-1} \left( 1 + (-\frac{x}{2} - \frac{x^2}{3} - \dots) + \frac{(-\frac{x}{2} - \dots)^2}{2} + \dots \right) = e^{-1} \left( 1 - \frac{x}{2} - \frac{x^2}{3} + \frac{x^2}{8} + \dots \right) = e^{-1} \left( 1 - \frac{x}{2} - \frac{5x^2}{24} + \dots \right)$.Substituting this into the limit:$L = \lim_{x \rightarrow 0^{+}} \frac{e^{-1} (1 - \frac{x}{2} - \frac{5x^2}{24} + \dots) - e^{-1}}{x^a} = e^{-1} \lim_{x \rightarrow 0^{+}} \frac{-\frac{x}{2} - \frac{5x^2}{24} - \dots}{x^a}$.For the limit to be a nonzero real number,the power of $x$ in the numerator must match the denominator $x^a$. Thus,$a = 1$.

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

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