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 31529 by abdo imad last updated on 09/Mar/18

$$find{lim}_{x\to 1^{-}}lnx.ln(1-x).$$

Commented by abdo imad last updated on 11/Mar/18

$${lim}_{x\to 1^{-}}lnx.ln(1-x)={lim}_{x\to 1^{-}}\frac{lnx}{x-1}(x-1)ln(1-x)but$$$${lim}_{x\to 1^{-}}\frac{lnx}{x-1}={ln}^{{}^{\prime }}\left(1\right)=1andch.1-x=tgive$$$${lim}_{x\to 1^{-}}(x-1)ln(1-x)={lim}_{t\to 0^{+}}-tln\left(t\right)=0\Rightarrow$$$${lim}_{x\to 1^{-}}lnx.ln(1-x)=0$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com