Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 96319 by  M±th+et+s last updated on 31/May/20

$$\underset{x\to 0}{lim}{\left(\frac{{(1+x)}^{\frac{1}{x}}}{e}\right)}^{\frac{1}{x}}$$

Answered by abdomathmax last updated on 31/May/20

$$\mathrm{f}\left(\mathrm{x}\right)={\left(\frac{{(1+\mathrm{x})}^{\frac{1}{\mathrm{x}}}}{\mathrm{e}}\right)}^{\frac{1}{\mathrm{x}}}\Rightarrow \ln \left(\mathrm{f}\left(\mathrm{x}\right)\right)=\frac{1}{\mathrm{x}}\ln \left(\frac{{(1+\mathrm{x})}^{\frac{1}{\mathrm{x}}}}{\mathrm{e}}\right)$$$$=\frac{1}{\mathrm{x}}(\ln {(1+\mathrm{x})}^{\frac{1}{\mathrm{x}}}-1)$$$$=\frac{1}{\mathrm{x}}(\frac{1}{\mathrm{x}}\ln (1+\mathrm{x})-1)=\frac{\ln (1+\mathrm{x})-\mathrm{x}}{{\mathrm{x}}^2}$$$$\mathrm{hodpiral}\to {\lim }_{\mathrm{x}\to 0}\frac{\ln (1+\mathrm{x})-\mathrm{x}}{{\mathrm{x}}^2}$$$$={\lim }_{\mathrm{x}\to 0}\frac{\frac{1}{1+\mathrm{x}}-1}{2\mathrm{x}}={\lim }_{\mathrm{x}\to 0}\frac{-1}{2(1+{\mathrm{x}}^2)}=-\frac{1}{2}\Rightarrow$$$$\ln \left(\mathrm{f}\left(\mathrm{x}\right)\right)\to -\frac{1}{2}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\to {\mathrm{e}}^{-\frac{1}{2}}=\frac{1}{\sqrt{\mathrm{e}}}$$

Commented by  M±th+et+s last updated on 31/May/20

$$niceworksirthankyou$$

Commented by mathmax by abdo last updated on 31/May/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com