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Question Number 194685 by cortano12 last updated on 13/Jul/23

Answered by qaz last updated on 14/Jul/23

$${(1+x)}^{\frac{1}{x}}=e^{\frac{1}{x}ln(1+x)}=e^{1-\frac{1}{2}x+\frac{1}{3}{x}^2+...}$$$$=e(1+(-\frac{1}{2}x+\frac{1}{3}{x}^2+...)+\frac{1}{2}{(-\frac{1}{2}x+\frac{1}{3}{x}^2+...)}^2+...$$$$=e(1-\frac{1}{2}x+\frac{11}{24}{x}^2+o\left(x^2\right))$$$$\Rightarrow \underset{x\to 0}{lim}\frac{{(1+x)}^{\frac{1}{x}}-e+\frac{1}{2}ex}{x^2}=\frac{11}{24}e$$

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