Question Number 173190 by mathlove last updated on 08/Jul/22
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}\left(\mathrm{2}−{e}^{{x}} \right)}{{x}+{lnx}}=? \\ $$
Answered by thfchristopher last updated on 08/Jul/22
$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{−{e}^{{x}} }{\mathrm{2}−{e}^{{x}} }}{\mathrm{1}+\frac{\mathrm{1}}{{x}}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−{xe}^{{x}} }{\left(\mathrm{2}−{e}^{{x}} \right)\left({x}+\mathrm{1}\right)} \\ $$$$=\mathrm{0} \\ $$
Answered by CElcedricjunior last updated on 08/Jul/22
$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{2}−\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \right)}{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{lnx}}}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$.........{le}\:{c}\acute {{e}l}\grave {{e}bre}\:{cedric}\:{junior}........ \\ $$