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Question Number 168055 by qaz last updated on 01/Apr/22

$$\mathrm{Calculate}::\underset{\mathrm{x}\to 0}{\lim }\frac{{(1+\mathrm{x})}^{-\frac{1}{{\mathrm{x}}^3}}}{\mathrm{x}}=?$$

Answered by LEKOUMA last updated on 03/Apr/22

$$\underset{x\to 0}{\lim }\frac{e^{((1+x)-1)\frac{1}{x^3}}}{x}=\underset{x\to 0}{\lim }\frac{e^{\frac{1}{x^2}}}{x}=+\mathrm{\infty }$$

Answered by Mathspace last updated on 04/Apr/22

$$f\left(x\right)=\frac{{(1+x)}^{-\frac{1}{x^3}}}{x}\Rightarrow$$$$f\left(x\right)=\frac{1}{x}{e}^{-\frac{1}{x^3}ln(1+x)}$$$${ln}^{\prime }(1+x)=\frac{1}{1+x}=1-x+o\left(x^2\right)$$$$\Rightarrow ln(1+x)\sim x\Rightarrow$$$$-\frac{1}{x^3}ln(1+x)\sim -\frac{1}{x^2}(x\to 0)\Rightarrow$$$$f\left(x\right)\sim \frac{1}{x}{e}^{-\frac{1}{x^2}}(\frac{1}{x}=y)$$$$(x\to o^{+}\Rightarrow y\to +\mathrm{\infty })\Rightarrow$$$$limf\left(x\right)={lim}_{y\to +\mathrm{\infty }}{ye}^{-y^2}=0$$$$$$

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