The value of mathop {lim }limits_{x to 0} fra | vedclass

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(A) We know that $(1 + x)^{1/x} = e^{\frac{1}{x} \ln(1 + x)}$.Using the expansion $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$,we get:$(1

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$\frac{11e}{24}$

Vedclass · Answer

(A) We know that $(1 + x)^{1/x} = e^{\frac{1}{x} \ln(1 + x)}$.Using the expansion $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$,we get:$(1 + x)^{1/x} = e^{\frac{1}{x}(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots)} = e^{1 - \frac{x}{2} + \frac{x^2}{3} - \dots} = e \cdot e^{-\frac{x}{2} + \frac{x^2}{3} - \dots}$.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}) + \frac{1}{2}(-\frac{x}{2})^2 + O(x^3) ight]$$= e \left[ 1 - \frac{x}{2} + \frac{x^2}{3} + \frac{x^2}{8} + O(x^3) ight] = e \left[ 1 - \frac{x}{2} + \frac{11x^2}{24} + O(x^3) ight]$.Substituting this into the limit:$\mathop {\lim }\limits_{x 	o 0} \frac{e(1 - \frac{x}{2} + \frac{11x^2}{24}) - e + \frac{ex}{2}}{x^2} = \mathop {\lim }\limits_{x 	o 0} \frac{e - \frac{ex}{2} + \frac{11ex^2}{24} - e + \frac{ex}{2}}{x^2} = \mathop {\lim }\limits_{x 	o 0} \frac{11ex^2}{24x^2} = \frac{11e}{24}$.

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/x}} - e + \frac{1}{2}ex}}{{{x^2}}}$ is

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