Question Number 105331 by Ar Brandon last updated on 27/Jul/20
![lim_(t→0) (1/t)ln[1−((ln(1+t))/(ln(t)))]](https://www.tinkutara.com/question/Q105331.png)
$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{t}}\mathrm{ln}\left[\mathrm{1}−\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{ln}\left(\mathrm{t}\right)}\right] \\ $$
Commented by bubugne last updated on 28/Jul/20
![= lim_(t→0) (1/t)ln[((ln(t)− ln(1+t))/(ln(t)))] = lim_(t→0) (1/t)ln[((ln((t/(t+1))))/(ln(t)))] = lim_(t→0) (1/t)ln[((ln(t))/(ln(t)))] = lim_(t→0) (1/t)ln(1)= 0](https://www.tinkutara.com/question/Q105363.png)
$$ \\ $$$$=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{t}}\mathrm{ln}\left[\frac{\mathrm{ln}\left(\mathrm{t}\right)−\:\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{ln}\left(\mathrm{t}\right)}\right] \\ $$$$=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{t}}\mathrm{ln}\left[\frac{\mathrm{ln}\left(\frac{\mathrm{t}}{\mathrm{t}+\mathrm{1}}\right)}{\mathrm{ln}\left(\mathrm{t}\right)}\right] \\ $$$$=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{t}}\mathrm{ln}\left[\frac{\mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{ln}\left(\mathrm{t}\right)}\right] \\ $$$$=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{t}}\mathrm{ln}\left(\mathrm{1}\right)=\:\mathrm{0} \\ $$
Commented by Dwaipayan Shikari last updated on 27/Jul/20
$$\mathrm{0} \\ $$
Commented by Ar Brandon last updated on 27/Jul/20
May I know how you got there, please ?