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 129086 by Ar Brandon last updated on 12/Jan/21

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

Answered by Dwaipayan Shikari last updated on 12/Jan/21

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

Commented by Ar Brandon last updated on 13/Jan/21

Cool bro ! ����

Answered by liberty last updated on 12/Jan/21

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

Commented by Ar Brandon last updated on 13/Jan/21

Same method I used. Thanks for affirmation.

Terms of Service

Privacy Policy

Contact: info@tinkutara.com