Limits Questions

Question Number 105331 by Ar Brandon last updated on 27/Jul/20

$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{t}}\mathrm{ln}\left[\mathrm{1}−\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{ln}\left(\mathrm{t}\right)}\right] \\$$

Commented by bubugne last updated on 28/Jul/20

$$\\$$$$=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{t}}\mathrm{ln}\left[\frac{\mathrm{ln}\left(\mathrm{t}\right)−\:\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{ln}\left(\mathrm{t}\right)}\right] \\$$$$=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{t}}\mathrm{ln}\left[\frac{\mathrm{ln}\left(\frac{\mathrm{t}}{\mathrm{t}+\mathrm{1}}\right)}{\mathrm{ln}\left(\mathrm{t}\right)}\right] \\$$$$=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{t}}\mathrm{ln}\left[\frac{\mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{ln}\left(\mathrm{t}\right)}\right] \\$$$$=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{t}}\mathrm{ln}\left(\mathrm{1}\right)=\:\mathrm{0} \\$$

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 ?