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 129655 by liberty last updated on 17/Jan/21

  lim_(x→∞) (log _x (2021x))^(log x)  =?                                                   ⌈∗⌉((√π^(ℓiberty) ) )

$$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{log}\:_{{x}} \left(\mathrm{2021}{x}\right)\right)^{\mathrm{log}\:{x}} \:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\lceil\ast\rceil\left(\sqrt{\pi^{\ell\mathrm{iberty}} }\:\right) \\ $$

Answered by bemath last updated on 17/Jan/21

 lim_(x→∞) (((ln (2021x))/(ln x)))^(ln x) = lim_(x→∞) (1+((ln 2021)/(ln x)))^(ln x)     lim_(x→∞) [(1+(1/((((ln x)/(ln 2021))))))((ln x)/(ln 2021)) ]^(ln 2021) = e^(ln 2021)  = 2021

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{ln}\:\left(\mathrm{2021x}\right)}{\mathrm{ln}\:\mathrm{x}}\right)^{\mathrm{ln}\:\mathrm{x}} =\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{ln}\:\mathrm{2021}}{\mathrm{ln}\:\mathrm{x}}\right)^{\mathrm{ln}\:\mathrm{x}} \\ $$$$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\left(\mathrm{1}+\frac{\mathrm{1}}{\left(\frac{\mathrm{ln}\:\mathrm{x}}{\mathrm{ln}\:\mathrm{2021}}\right)}\right)\frac{\mathrm{ln}\:\mathrm{x}}{\mathrm{ln}\:\mathrm{2021}}\:\right]^{\mathrm{ln}\:\mathrm{2021}} =\:\mathrm{e}^{\mathrm{ln}\:\mathrm{2021}} \:=\:\mathrm{2021} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com