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This is a question given in our weekly test.

$$f = \lim_{x\to 0^+}⁡\{[(1+x)^{1/x}]/e\}^{1/x}.$$

Find the value of $f$. I tried to use 1^ infinity form but I didn't get it. So anybody please help me.

sai saandeep
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3 Answers3

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Your limit $f$ exists if and only if its logarithm exists: \begin{align} \log f &=\lim_{x\to0^+}\log\bigl(\bigl((1+x)^{1/x})/e\bigr)^{1/x}\bigr) \\[6px] &=\lim_{x\to0^+}\frac{\dfrac{1}{x}\log(1+x)-1}{x} \\[6px] &=\lim_{x\to0^+}\frac{\log(1+x)-x}{x^2} \\[6px] &=\lim_{x\to0^+}\frac{x-x^2/2+o(x^2)-x}{x^2}=-\frac{1}{2} \end{align} If you don't trust Taylor expansions (or cannot use them), $$ \lim_{x\to0^+}\frac{\log(1+x)-x}{x^2} \overset{*}{=} \lim_{x\to0^+}\frac{\dfrac{1}{1+x}-1}{2x} = \lim_{x\to0^+}\frac{-x}{2x(1+x)} $$ (where $\overset{*}{=}$ denotes an application of l'Hôpital).

egreg
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\begin{align} f&=\lim_{x\to0}\left(\frac{(1+x)^{1/x}}e\right)^{1/x}\\ &=\lim_{x\to0}\left(\frac1{e^x}+\frac{x}{e^x}\right)^{1/x^2}\\ &=\lim_{x\to0}\left(1+\frac{1+x-e^x}{e^x}\right)^{1/x^2}\\ &=\lim_{x\to0}\left(\left(1+\frac{1+x-e^x}{e^x}\right)^{\cfrac{e^x}{1+x-e^x}}\right)^{\cfrac{1+x-e^x}{x^2e^x}}\\ &=\lim_{x\to0}\quad e^{\cfrac{1+x-e^x}{x^2}\cfrac1{e^x}}\\ &=e^{\ \lim_{x\to0}\cfrac{1+x-e^x}{x^2}\cfrac1{e^x}}\\ &=e^{\ \lim_{x\to0}\cfrac{1+x-e^x}{x^2}}\\ &\overset{\text{L'Hopital}}{=}e^{\ \lim_{x\to0}\cfrac{1-e^x}{2x}}\\ &\overset{\text{L'Hopital}}{=}e^{\ \lim_{x\to0}\cfrac{-e^x}{2}}\\ &=e^{-1/2} \end{align}

Kay K.
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$$\lim\limits_{x\to 0^{+}}\left(\frac{(1+x)^{1/x}}{x}\right)^{1/x}=\lim\limits_{x\to 0^{+}}e^{-1/x}\left((x+1)^{1/x}\right)^{1/x}=\lim\limits_{x\to 0^{+}}\exp\left(\frac{\ln((x+1)^{1/x})}{x}-\frac 1 x\right)$$

$$=\exp\left(\lim\limits_{x\to 0^{+}}\left(\frac{\ln\left((1+x)^{1/x}\right)}{x}-\frac 1 x\right)\right)^{1/x}=\exp\left(\lim\limits_{x\to 0^{+}}\left(\frac{\frac{\ln(x+1)}{x}}{x}-\frac 1 x\right)\right)\\ =\exp\left(\lim\limits_{x\to 0^{+}}\frac{\ln(x+1)-x}{x^2}\right)$$

$$\overset{\text{L'Hopital}}{=}\exp \left(\lim\limits_{x\to 0^{+}}\underbrace{-\frac{1}{2(x+1)}}_{\to -1/2}\right)=e^{-1/2}$$

3SAT
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