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Question Number 80417 by jagoll last updated on 03/Feb/20

$$\underset{x\to 0}{\lim }\frac{{(1+x)}^{\frac{1}{x}}-e}{x}=?$$

Commented by MJS last updated on 03/Feb/20

$$\mathrm{I}\mathrm{tried}\mathrm{to}\mathrm{approximate}\mathrm{and}\mathrm{got}\mathrm{something}$$$$\mathrm{very}\mathrm{close}\mathrm{to}-\frac{\mathrm{e}}{2}\mathrm{but}\mathrm{I}\mathrm{cannot}\mathrm{prove}\mathrm{the}$$$$\mathrm{exact}\mathrm{value}$$

Commented by mr W last updated on 03/Feb/20

$${(1+x)}^{\frac{1}{x}}=e(1-\frac{x}{2}+\frac{11x^2}{24}-...)$$$${(1+x)}^{\frac{1}{x}}-e=e(-\frac{x}{2}+\frac{11x^2}{24}-...)$$$$\frac{{(1+x)}^{\frac{1}{x}}-e}{x}=e(-\frac{1}{2}+\frac{11x}{24}-...)\to -\frac{e}{2}$$

Commented by jagoll last updated on 03/Feb/20

$$viaTaylorseriessir?$$

Commented by mr W last updated on 03/Feb/20

$$yes!$$

Commented by jagoll last updated on 03/Feb/20

$$thankyousir$$

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