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Question Number 162925 by qaz last updated on 02/Jan/22

$$\underset{\mathrm{x}\to 0}{\lim }\frac{1}{{\mathrm{x}}^4}[{(1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{3})}^{\frac{1}{\mathrm{x}+\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{3}}}-{(1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^4}{4})}^{\frac{1}{\mathrm{x}+\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^4}{4}}}]=?$$

Answered by Ar Brandon last updated on 02/Jan/22

$$=\underset{x\to 0}{\lim }\{e^{\left(\frac{1}{x+\frac{x^2}{2}+\frac{x^3}{3}}\right)\ln (1+x+\frac{x^2}{2}+\frac{x^3}{3})}-e^{\left(\frac{1}{x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}}\right)\ln (1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4})}\}$$$$=\underset{x\to 0}{\lim }\{e^{\frac{1}{x+\frac{x^2}{2}+\frac{x^3}{3}}(x+\frac{x^2}{2}+\frac{x^3}{3})}-e^{\frac{1}{x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}}(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4})}\}$$$$=e-e=0$$

Commented by qaz last updated on 02/Jan/22

$${\mathrm{what}}^{\prime }\mathrm{s}\mathrm{about}\mathrm{Q}162926?$$

Commented by qaz last updated on 02/Jan/22

$$\mathrm{sorry}.\mathrm{I}\mathrm{typed}\mathrm{wrong}$$

Commented by Ar Brandon last updated on 02/Jan/22

$$\frac{e}{8}$$

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