mathop {lim }limits_{x to 0} frac{{{{(1 + x)} | vedclass

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Question

(A) Let $f(x) = (1 + x)^{1/x}$.Using the expansion $\log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$,we have:$(1 + x)^{1/x} = e^{\frac{1}{x} \

Vedclass · Accepted Answer

$-e/2$

Vedclass · Answer

(A) Let $f(x) = (1 + x)^{1/x}$.Using the expansion $\log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$,we have:$(1 + x)^{1/x} = e^{\frac{1}{x} \log(1 + x)} = e^{\frac{1}{x} (x - \frac{x^2}{2} + \frac{x^3}{3} - \dots)} = e^{1 - \frac{x}{2} + \frac{x^2}{3} - \dots} = e \cdot e^{-\frac{x}{2} + \frac{x^2}{3} - \dots}$.Using the expansion $e^u = 1 + u + \frac{u^2}{2!} + \dots$,where $u = -\frac{x}{2} + \frac{x^2}{3} - \dots$:$(1 + x)^{1/x} = e \left( 1 + (-\frac{x}{2} + \frac{x^2}{3} - \dots) + \frac{(-\frac{x}{2} + \dots)^2}{2} + \dots ight) = e - \frac{ex}{2} + O(x^2)$.Thus,$\mathop {\lim }\limits_{x 	o 0} \frac{(1 + x)^{1/x} - e}{x} = \mathop {\lim }\limits_{x 	o 0} \frac{(e - \frac{ex}{2} + O(x^2)) - e}{x} = \mathop {\lim }\limits_{x 	o 0} (-\frac{e}{2} + O(x)) = -\frac{e}{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/x}} - e}}{x}$ equals

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