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 72495 by Tony Lin last updated on 29/Oct/19

$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{lnx}{x}\right)}^{\frac{lnx}{x}}=?$$

Commented by mathmax by abdo last updated on 29/Oct/19

$$letf\left(x\right)={\left(\frac{lnx}{x}\right)}^{\frac{lnx}{x}}vhangementlnx=tgive$$$$f\left(x\right)=g\left(t\right)={\left(\frac{t}{e^t}\right)}^{\frac{t}{e^t}}={\left({te}^{-t}\right)}^{{te}^{-t}}x\to +\mathrm{\infty }\iff t\to +\mathrm{\infty }$$$$g\left(t\right)=e^{{te}^{-t}ln\left(t{e}^{-t}\right)}=e^{t{e}^{-t}(lnt-t)}=e^{tln\left(t\right)e^{-t}}{e}^{-t^2{e}^{-t}}$$$$but{lim}_{t\to +\mathrm{\infty }}tln\left(t\right)e^{-t}=0and{lim}_{t\to +\mathrm{\infty }}{e}^{-t^2{e}^{-t}}=1\Rightarrow$$$${lim}_{x\to +\mathrm{\infty }}f\left(x\right)=1$$

Commented by Tony Lin last updated on 30/Oct/19

$$thankssir$$

Commented by mathmax by abdo last updated on 30/Oct/19

$$youarewelcome.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com